Explanation:
Okay, so I need to simplify this big fraction where the numerator is the product of terms like (2^4 + 1/4), (4^4 + 1/4), all the way up to (2n)^4 + 1/4, and the denominator is the product of similar terms but with odd numbers: (1^4 + 1/4), (3^4 + 1/4), ..., up to (2n - 1)^4 + 1/4.

First, let me write down the expression clearly to make sure I understand it:

$$
\frac{(2^4 + \frac{1}{4})(4^4 + \frac{1}{4}) \dotsm [(2n)^4 + \frac{1}{4}]}{(1^4 + \frac{1}{4})(3^4 + \frac{1}{4}) \dotsm [(2n - 1)^4 + \frac{1}{4}]}
$$

So, the numerator has terms for even numbers from 2 up to 2n, and the denominator has terms for odd numbers from 1 up to 2n - 1. Each term in both the numerator and denominator is of the form k^4 + 1/4, where k is an integer. My job is to simplify this entire expression.

Hmm, I remember that sometimes expressions like a^4 + 1/4 can be factored. Maybe I can factor each term in the numerator and denominator, and then see if there's some cancellation. Let me try that.

First, let me recall how to factor expressions like a^4 + b^4 or similar. Wait, here it's a^4 + 1/4. Let me think. Maybe I can write it as a^4 + (1/2)^2? But that's not a standard identity. Alternatively, perhaps using complex numbers? But that might complicate things. Alternatively, maybe I can use the Sophie Germain identity or something similar?

Wait, Sophie Germain identity is for a^4 + 4b^4, right? Which factors into (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2). But in our case, we have a^4 + (1/4). Let me check if this can be related to Sophie Germain's identity.

Let me try to manipulate the expression. Let me denote k^4 + 1/4. Let me write 1/4 as (1/2)^2. So, we have k^4 + (1/2)^2. Hmm, which is not directly a Sophie Germain identity. Maybe if I adjust it somehow?

Alternatively, maybe I can think of it as k^4 + 4*(1/4)^2? Wait, 4*(1/4)^2 is 4*(1/16) = 1/4. So, that's exactly the same. So, k^4 + 4*(1/4)^2. So then, using Sophie Germain identity, which is a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2).

So here, if we take a = k and b = 1/2, then:

k^4 + 4*(1/2)^4 = k^4 + 4*(1/16) = k^4 + 1/4. Wait, but 4*(1/2)^4 is 4*(1/16) = 1/4. So, indeed, the expression k^4 + 1/4 can be written as a^4 + 4b^4 where a = k and b = 1/2. Therefore, we can factor it as:

(k^2 + 2*(k)*(1/2) + 2*(1/2)^2)(k^2 - 2*(k)*(1/2) + 2*(1/2)^2)

Simplify each term:

First factor:

k^2 + 2*(k)*(1/2) + 2*(1/2)^2 = k^2 + k + 2*(1/4) = k^2 + k + 1/2

Second factor:

k^2 - 2*(k)*(1/2) + 2*(1/2)^2 = k^2 - k + 1/2

Therefore, we have:

k^4 + 1/4 = (k^2 + k + 1/2)(k^2 - k + 1/2)

So that's a useful factorization!

Therefore, each term in the numerator and denominator can be factored into two quadratic terms. Maybe this will allow for cancellation?

Let me write down that factorization:

For each term (k^4 + 1/4) = (k^2 + k + 1/2)(k^2 - k + 1/2).

So then, the entire numerator is the product over even k from 2 to 2n of (k^2 + k + 1/2)(k^2 - k + 1/2), and the denominator is the product over odd k from 1 to 2n - 1 of (k^2 + k + 1/2)(k^2 - k + 1/2). Therefore, if I write the entire fraction as:

[Product_{even k} (k^2 + k + 1/2)(k^2 - k + 1/2)] / [Product_{odd k} (k^2 + k + 1/2)(k^2 - k + 1/2)]

Which can be written as:

[Product_{even k} (k^2 + k + 1/2) * Product_{even k} (k^2 - k + 1/2)] / [Product_{odd k} (k^2 + k + 1/2) * Product_{odd k} (k^2 - k + 1/2)]

So maybe splitting the numerator and denominator into two separate products. Let me think about how to handle each of these. Let me consider the products of (k^2 + k + 1/2) and (k^2 - k + 1/2) separately for even and odd k.

Alternatively, perhaps there's some telescoping happening here. Let me try to see if the terms can be rearranged or related to each other.

First, let me note that for a general k, if I consider (k^2 + k + 1/2) and (k^2 - k + 1/2), these are similar to quadratic expressions. Maybe if we write them in terms of (k + something), or relate them to terms for other values of k?

Alternatively, let me note that for each term in the numerator and denominator, after factoring, they have these quadratic factors. Let me check for specific values of k. Maybe I can compute a few terms to see a pattern.

Let me take k as even numbers first. Let me pick k = 2:

For k=2: (2^2 + 2 + 1/2) = 4 + 2 + 0.5 = 6.5 = 13/2

(2^2 - 2 + 1/2) = 4 - 2 + 0.5 = 2.5 = 5/2

Similarly, for k=4:

(4^2 + 4 + 1/2) = 16 + 4 + 0.5 = 20.5 = 41/2

(4^2 - 4 + 1/2) = 16 - 4 + 0.5 = 12.5 = 25/2

Similarly, for odd k, say k=1:

(1^2 + 1 + 1/2) = 1 + 1 + 0.5 = 2.5 = 5/2

(1^2 - 1 + 1/2) = 1 - 1 + 0.5 = 0.5 = 1/2

For k=3:

(9 + 3 + 0.5) = 12.5 = 25/2

(9 - 3 + 0.5) = 6.5 = 13/2

Hmm, interesting. For k=2, (k^2 + k + 1/2) is 13/2 and (k^2 - k + 1/2) is 5/2. For k=1, which is odd, (k^2 + k + 1/2) is 5/2 and (k^2 - k + 1/2) is 1/2. Similarly, for k=3, odd, the first term is 25/2 and the second term is 13/2. Wait, for k=2 even, (k^2 + k + 1/2) is 13/2, and for k=3 odd, the (k^2 - k + 1/2) term is 13/2. So there's a connection here. Similarly, for k=4 even, the first term is 41/2, which is same as k=5 odd's (k^2 - k + 1/2) term? Let me check k=5:

(5^2 - 5 + 1/2) = 25 - 5 + 0.5 = 20.5 = 41/2. Yes! Exactly. Similarly, for k=1 odd, (k^2 - k + 1/2) = 1/2, which is same as for k=0? Wait, k=0? If I plug k=0 into (k^2 + k + 1/2) it would be 0 + 0 + 1/2 = 1/2. But k=0 is not in our original products. However, this suggests that the terms for even k in the numerator's first factor (k^2 + k + 1/2) are equal to the terms for odd k in the denominator's second factor (k^2 - k + 1/2). Similarly, for the other way around?

Wait, let me think. Let me consider the general case. Suppose I have an even number k = 2m. Then:

(k^2 + k + 1/2) = (4m² + 2m + 1/2)

Similarly, for the odd number k = 2m - 1:

(k^2 - k + 1/2) = [(2m - 1)^2 - (2m - 1) + 1/2]

Let me compute that:

(4m² - 4m + 1) - (2m - 1) + 1/2 = 4m² - 4m + 1 - 2m + 1 + 1/2 = 4m² - 6m + 2 + 1/2 = 4m² - 6m + 5/2

Wait, but how does that compare to the even term?

Wait, earlier for k=2 (m=1):

k^2 + k + 1/2 = 4 + 2 + 0.5 = 6.5 = 13/2. For the odd k=1 (m=1), (k^2 - k + 1/2) = 1 - 1 + 0.5 = 0.5 = 1/2. Wait, that doesn't seem to match. Wait earlier I thought there was a connection, but maybe my previous thought was incorrect. Wait, let me check again with k=2 (even) and k=3 (odd):

For k=2 (even): (k^2 + k + 1/2) = 4 + 2 + 0.5 = 6.5 = 13/2

For k=3 (odd): (k^2 - k + 1/2) = 9 - 3 + 0.5 = 6.5 = 13/2. Ah! So here, for k=2 (even) and k=3 (odd), the first factor of the even term is equal to the second factor of the odd term. Similarly, for k=4 (even), (k^2 + k + 1/2) = 16 + 4 + 0.5 = 20.5 = 41/2. For k=5 (odd), (k^2 - k + 1/2) = 25 - 5 + 0.5 = 20.5 = 41/2. So same here. Similarly, for k=1 (odd), (k^2 - k + 1/2) = 1 - 1 + 0.5 = 0.5 = 1/2. While for k=0 (even), (k^2 + k + 1/2) = 0 + 0 + 0.5 = 0.5 = 1/2. But k=0 isn't in our denominator.

So the idea is that for each even number k = 2m, the term (k^2 + k + 1/2) is equal to the term (k^2 - k + 1/2) for the next odd number, which is k = 2m + 1. Wait, but in the case of k=2 (even), which is m=1, the term is (k^2 + k + 1/2) = 13/2, which equals (k^2 - k + 1/2) for k=3 (odd). Similarly, for k=4 (even), m=2, the term (k^2 + k + 1/2) = 41/2 equals the term (k^2 - k + 1/2) for k=5 (odd).

Therefore, this seems to suggest that if we have the numerator terms (even k) with the first factor (k^2 + k + 1/2) and denominator terms (odd k) with the second factor (k^2 - k + 1/2), they might cancel out. Similarly, the second factor of the even k terms (k^2 - k + 1/2) would correspond to the first factor of the odd k terms?

Wait, let me check for k=2 (even) second factor: (k^2 - k + 1/2) = 4 - 2 + 0.5 = 2.5 = 5/2. Which is the same as for k=1 (odd) first factor (k^2 + k + 1/2) = 1 + 1 + 0.5 = 2.5 = 5/2. Similarly, for k=4 (even) second factor: 16 - 4 + 0.5 = 12.5 = 25/2, which is same as k=3 (odd) first factor: 9 + 3 + 0.5 = 12.5 = 25/2.

Therefore, this suggests that for each even term in the numerator, the first factor (k^2 + k + 1/2) is equal to the second factor of the next odd term in the denominator, and the second factor of the even term (k^2 - k + 1/2) is equal to the first factor of the previous odd term in the denominator.

So if we consider the entire products, there might be cancellation between numerator and denominator. Let me try to write this out more formally.

Let me denote:

For each even k = 2m (where m ranges from 1 to n), we have:

Term in numerator: (k^2 + k + 1/2)(k^2 - k + 1/2)

For each odd k = 2m - 1 (where m ranges from 1 to n), we have:

Term in denominator: (k^2 + k + 1/2)(k^2 - k + 1/2)

So the entire fraction is:

[Numerator: product_{m=1}^n ( ( (2m)^2 + 2m + 1/2 ) * ( (2m)^2 - 2m + 1/2 ) ) ] / [Denominator: product_{m=1}^n ( ( (2m - 1)^2 + (2m - 1) + 1/2 ) * ( (2m - 1)^2 - (2m - 1) + 1/2 ) ) ]

But as we saw, ( (2m)^2 + 2m + 1/2 ) equals ( (2m + 1)^2 - (2m + 1) + 1/2 ), since if we take k = 2m + 1 (which is odd), then k^2 - k + 1/2 = (2m + 1)^2 - (2m + 1) + 1/2 = 4m² + 4m + 1 - 2m - 1 + 1/2 = 4m² + 2m + 0 + 1/2 = 4m² + 2m + 1/2. Which is exactly equal to (2m)^2 + 2m + 1/2, since (2m)^2 is 4m².

Similarly, ( (2m)^2 - 2m + 1/2 ) = (2m - 1)^2 + (2m - 1) + 1/2. Let me check:

k = 2m - 1 (odd). Then k^2 + k + 1/2 = (2m - 1)^2 + (2m - 1) + 1/2 = 4m² - 4m + 1 + 2m - 1 + 1/2 = 4m² - 2m + 0 + 1/2 = 4m² - 2m + 1/2. Which is equal to (2m)^2 - 2m + 1/2.

Therefore, this means that for each even term in the numerator, the first factor (k^2 + k + 1/2) is equal to the second factor of the next odd term in the denominator. Similarly, the second factor of the even term (k^2 - k + 1/2) is equal to the first factor of the previous odd term in the denominator.

Therefore, when we take the entire product, these terms will cancel out with the denominator. Let me think about how the products would look.

Let me write the numerator's factors and denominator's factors:

Numerator has for each even k=2m:

Factor1: ( (2m)^2 + 2m + 1/2 )

Factor2: ( (2m)^2 - 2m + 1/2 )

Denominator has for each odd k=2m-1:

Factor1: ( (2m - 1)^2 + (2m - 1) + 1/2 )

Factor2: ( (2m - 1)^2 - (2m - 1) + 1/2 )

But as established, Factor1 of numerator (even) = Factor2 of denominator (next odd). Similarly, Factor2 of numerator (even) = Factor1 of denominator (previous odd).

Therefore, if we look at the entire product, when we take the numerator over the denominator, all these terms will telescope. Let's think about how.

Let me denote:

For the numerator:

Product over m=1 to n of [Factor1_even(m) * Factor2_even(m)]

For the denominator:

Product over m=1 to n of [Factor1_odd(m) * Factor2_odd(m)]

But since Factor1_even(m) = Factor2_odd(m+1), and Factor2_even(m) = Factor1_odd(m). Wait, let me check:

For Factor1_even(m) = (2m)^2 + 2m + 1/2. Which is equal to (2m + 1)^2 - (2m + 1) + 1/2. Which is the Factor2_odd(m+1).

Similarly, Factor2_even(m) = (2m)^2 - 2m + 1/2. Which is equal to (2m - 1)^2 + (2m - 1) + 1/2. Which is the Factor1_odd(m).

Therefore, if we consider the entire products:

Numerator: [Product_{m=1}^n (Factor1_even(m) * Factor2_even(m))]

Denominator: [Product_{m=1}^n (Factor1_odd(m) * Factor2_odd(m))]

But Factor1_even(m) = Factor2_odd(m+1) and Factor2_even(m) = Factor1_odd(m). Therefore, substituting into the numerator:

Product_{m=1}^n [Factor2_odd(m+1) * Factor1_odd(m)]

Which is equal to [Product_{m=1}^n Factor1_odd(m) * Product_{m=1}^n Factor2_odd(m+1)]

But Product_{m=1}^n Factor2_odd(m+1) = Product_{m=2}^{n+1} Factor2_odd(m)

Similarly, Product_{m=1}^n Factor1_odd(m) = Product_{m=1}^n Factor1_odd(m)

Therefore, the entire numerator becomes:

[Product_{m=1}^n Factor1_odd(m)] * [Product_{m=2}^{n+1} Factor2_odd(m)]

So when we divide by the denominator, which is [Product_{m=1}^n Factor1_odd(m) * Factor2_odd(m)], we get:

[Product_{m=1}^n Factor1_odd(m) * Product_{m=2}^{n+1} Factor2_odd(m)] / [Product_{m=1}^n Factor1_odd(m) * Product_{m=1}^n Factor2_odd(m)]

This simplifies to [Product_{m=2}^{n+1} Factor2_odd(m)] / [Product_{m=1}^n Factor2_odd(m)]

Which is equal to [Product_{m=2}^{n+1} Factor2_odd(m)] / [Product_{m=1}^n Factor2_odd(m)]

This simplifies further to [Factor2_odd(n+1) / Factor2_odd(1)]

Because when you divide the two products, the terms from m=2 to n cancel out, leaving Factor2_odd(n+1) divided by Factor2_odd(1).

So, the entire expression simplifies to Factor2_odd(n+1) divided by Factor2_odd(1).

So, what is Factor2_odd(m)? Recall that Factor2_odd(m) is (k^2 - k + 1/2) where k = 2m - 1.

Therefore, Factor2_odd(n+1) is [(2(n+1) - 1)^2 - (2(n+1) - 1) + 1/2] = [(2n + 1)^2 - (2n + 1) + 1/2]

Similarly, Factor2_odd(1) is [(2*1 - 1)^2 - (2*1 - 1) + 1/2] = [1^2 - 1 + 1/2] = 0 + 1/2 = 1/2

So let me compute Factor2_odd(n+1):

Compute [(2n + 1)^2 - (2n + 1) + 1/2]

First expand (2n + 1)^2 = 4n² + 4n + 1

Subtract (2n + 1): 4n² + 4n + 1 - 2n - 1 = 4n² + 2n

Add 1/2: 4n² + 2n + 1/2

So Factor2_odd(n+1) = 4n² + 2n + 1/2

Therefore, the entire expression simplifies to [4n² + 2n + 1/2] / [1/2]

Which is equal to [ (4n² + 2n + 1/2) ] / (1/2) = 2*(4n² + 2n + 1/2) = 8n² + 4n + 1

Wait, let me check that arithmetic again.

Wait, if you have (a) / (1/2) = 2a. So:

[4n² + 2n + 1/2] divided by 1/2 is equal to 2*(4n² + 2n + 1/2) = 8n² + 4n + 1.

Yes. So the entire expression simplifies to 8n² + 4n + 1.

But let me verify this with a small value of n to check if this is correct. Let's take n=1.

Case n=1:

Numerator: (2^4 + 1/4) = 16 + 1/4 = 65/4

Denominator: (1^4 + 1/4) = 1 + 1/4 = 5/4

So the entire expression is (65/4) / (5/4) = 65/5 = 13.

Now according to the formula 8n² + 4n + 1 with n=1: 8(1) + 4(1) + 1 = 8 + 4 + 1 = 13. Correct.

Another test case: n=2.

Numerator terms: (2^4 + 1/4)(4^4 + 1/4) = (16 + 0.25)(256 + 0.25) = (16.25)(256.25)

Denominator terms: (1^4 + 1/4)(3^4 + 1/4) = (1.25)(81 + 0.25) = 1.25 * 81.25

Compute numerator: 16.25 * 256.25

Let me compute that:

But maybe it's easier to use fractions.

Original terms:

For n=2:

Numerator: (2^4 + 1/4) = 16 + 1/4 = 65/4

(4^4 + 1/4) = 256 + 1/4 = 1025/4

So numerator product: (65/4) * (1025/4) = (65 * 1025)/16

Denominator: (1^4 + 1/4) = 5/4; (3^4 + 1/4) = 81 + 1/4 = 325/4. So denominator product: (5/4)*(325/4) = (5*325)/16

Therefore, the entire expression is [(65*1025)/16] / [(5*325)/16] = (65*1025)/(5*325)

Simplify:

Divide numerator and denominator by 5: (13*1025)/(325)

Note that 325 = 25*13, so:

(13*1025)/(25*13) = 1025/25 = 41.

Now according to the formula 8n² + 4n + 1 with n=2: 8*(4) + 4*2 + 1 = 32 + 8 + 1 = 41. Correct again!

Another check with n=3.

Compute formula: 8*(9) + 4*3 + 1 = 72 + 12 + 1 = 85.

Let me compute the original expression for n=3:

Numerator: (2^4 + 1/4)(4^4 + 1/4)(6^4 + 1/4) = (65/4)(1025/4)(1296 + 1/4) = (65/4)(1025/4)(5185/4)

Denominator: (1^4 + 1/4)(3^4 + 1/4)(5^4 + 1/4) = (5/4)(325/4)(625 + 1/4) = (5/4)(325/4)(2501/4)

Compute the entire expression:

[ (65/4)(1025/4)(5185/4) ] / [ (5/4)(325/4)(2501/4) ]

This simplifies to [65 * 1025 * 5185] / [5 * 325 * 2501]

Let me compute step by step:

First, 65 / 5 = 13

Then, 1025 / 325 = 1025 / 325. Let me divide numerator and denominator by 25: 41 / 13.

Then, 5185 / 2501. Let me check if 2501 * 2 = 5002, so 5185 - 5002 = 183, so maybe not. Alternatively, let me check 2501: 2501 is 50^2 + 1 = 2500 + 1 = 2501. Wait, and 5185 is 5185. Let me check if 2501 divides into 5185.

Wait, 2501 * 2 = 5002, 5185 - 5002 = 183. So no. Wait, but perhaps there's a relationship here. Let me recall that earlier, our telescoping led us to 8n² + 4n + 1. For n=3, it's 85. So the result should be 85. Let me check if [13 * (41/13) * (5185 / 2501)] equals 85.

Simplify 13 * (41/13) = 41. Then 41 * (5185 / 2501) = ?

Let me check 5185 divided by 2501. Let me compute 2501 * 2 = 5002. 5185 - 5002 = 183. So 5185 = 2501 * 2 + 183. Let me check if 183 relates to something. Wait, but if the result is supposed to be 85, then 41 * (5185 / 2501) = 85 implies 5185 / 2501 = 85 / 41. Let me compute 85 / 41 ≈ 2.073. Let me compute 5185 / 2501 ≈ 5185 / 2501 ≈ 2.073. Let me check 2501 * 85 = 2501 * 80 + 2501 * 5 = 200,080 + 12,505 = 212,585. And 41 * 5185 = 41 * 5000 + 41 * 185 = 205,000 + 7,585 = 212,585. Yes! Therefore, 41 * (5185 / 2501) = 85. Therefore, the total expression is indeed 85. Which matches the formula.

Therefore, the formula seems correct. Therefore, the simplified expression is 8n² + 4n + 1.

Wait, but let me check with n=1, 2, 3 and it works, so seems solid.

But let me also check if there is another way to write 8n² + 4n + 1. Maybe factoring?

8n² + 4n + 1. Let me see:

It doesn't factor nicely with integer coefficients. Let me check discriminant: 16 - 32 = -16? Wait no. Wait discriminant is b² - 4ac = 16 - 32 = -16? Wait no, 8n² + 4n + 1. Discriminant is 16 - 32 = -16. So it doesn't factor over reals? Wait, no, discriminant is 16 - 4*8*1 = 16 - 32 = -16. Therefore, complex roots. Therefore, the expression is irreducible over real numbers. Therefore, the simplified form is 8n² + 4n + 1.

But let me check again if the telescoping was correctly done. The key step was realizing that the numerator factors cancel with denominator factors, leaving only the last term in numerator and the first term in denominator? Wait, earlier we had that the entire expression simplified to Factor2_odd(n+1)/Factor2_odd(1). And Factor2_odd(n+1) was 4n² + 2n + 1/2, then divided by 1/2 gives 8n² + 4n + 1.

Yes. So that seems correct. Therefore, the simplified expression is 8n² + 4n + 1.

But let me check if there is another way to write this. For example, note that 8n² + 4n + 1 can be written as (2n + 1)^2 + (2n)^2? Let me check:

(2n + 1)^2 = 4n² + 4n + 1, and (2n)^2 = 4n². So sum is 8n² + 4n + 1. Yes. So 8n² + 4n + 1 = (2n)^2 + (2n + 1)^2. Interesting. But maybe that's not necessary.

Alternatively, maybe written as 4*(2n² + n) + 1, but again not sure. Anyway, the answer is 8n² + 4n + 1.

But let me check for n=1: 8 + 4 + 1 = 13, which matches. n=2: 32 + 8 + 1 = 41, which matches. n=3: 72 + 12 + 1 = 85. Correct.

Alternatively, maybe there's a different expression? Let me think again.

Wait, let me think back to the original factors.

We had the entire expression simplified to [Factor2_odd(n+1)] / [Factor2_odd(1)].

Factor2_odd(n+1) was [(2n + 1)^2 - (2n + 1) + 1/2] = 4n² + 2n + 1/2.

But if I compute 4n² + 2n + 1/2 divided by 1/2, it's 8n² + 4n + 1.

Alternatively, maybe there's a different way to express this? For example, if I consider that the entire original expression was telescoped to a single term, perhaps related to (some term with n)?

But given that the test cases work, and the derivation seems correct, I think that 8n² + 4n + 1 is the correct simplified expression. Therefore, the answer is \boxed{8n^2 + 4n + 1}.

But wait, let me check once more for n=1,2,3. For n=1: 8(1)^2 + 4(1) + 1 = 13, which matched. For n=2: 8*4 + 8 +1 = 32 + 8 + 1 = 41, correct. For n=3: 8*9 + 12 + 1 = 72 + 12 + 1 = 85. Correct.

Alternatively, is there a possibility that this expression can be simplified more? Let me check if 8n² + 4n + 1 can be written as (something else). For example, maybe related to (2n + 1)^2 + something?

(2n + 1)^2 = 4n² + 4n + 1. So 8n² + 4n + 1 = 2*(4n² + 2n) + 1. Hmm, not sure. Alternatively, maybe 8n² + 4n + 1 = 4n² + (4n² + 4n + 1) = 4n² + (2n + 1)^2. Which is another way to write it. But I think the problem just wants the simplified expression, and 8n² + 4n + 1 is a valid simplified form.

Therefore, I think that the answer is 8n² + 4n + 1.

**Final Answer**
\boxed{8n^2 + 4n + 1}

To simplify the given expression:

$$
\frac{(2^4 + \frac{1}{4})(4^4 + \frac{1}{4}) \dotsm [(2n)^4 + \frac{1}{4}]}{(1^4 + \frac{1}{4})(3^4 + \frac{1}{4}) \dotsm [(2n - 1)^4 + \frac{1}{4}]}
$$

we observe that each term in the numerator is of the form $(k^4 + \tfrac{1}{4})$ where $k = 2, 4, 6, \ldots, 2n$, and each term in the denominator is of the form $(k^4 + \tfrac{1}{4})$ where $k = 1, 3, 5, \ldots, 2n - 1$.

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### Step 1: Factorization using Sophie Germain Identity

We use the identity:

$$
a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)
$$

Let us apply this identity to $k^4 + \tfrac{1}{4}$ by writing it as:

$$
k^4 + \frac{1}{4} = k^4 + 4\left(\frac{1}{4}\right)^2
$$

So let $a = k$, $b = \frac{1}{2}$, then:

$$
k^4 + \frac{1}{4} = (k^2 + 2k \cdot \frac{1}{2} + 2 \cdot \left(\frac{1}{2}\right)^2)(k^2 - 2k \cdot \frac{1}{2} + 2 \cdot \left(\frac{1}{2}\right)^2)
$$

Simplifying:

$$
k^4 + \frac{1}{4} = (k^2 + k + \tfrac{1}{2})(k^2 - k + \tfrac{1}{2})
$$

Thus, every factor in both the numerator and denominator can be written as a product of these two quadratic expressions.

---

### Step 2: Telescoping Structure

Now write the entire numerator and denominator using the factorization:

- **Numerator:**

$$
\prod_{k=1}^n ( (2k)^2 + 2k + \tfrac{1}{2} ) \cdot ( (2k)^2 - 2k + \tfrac{1}{2} )
$$

- **Denominator:**

$$
\prod_{k=1}^n ( (2k - 1)^2 + (2k - 1) + \tfrac{1}{2} ) \cdot ( (2k - 1)^2 - (2k - 1) + \tfrac{1}{2} )
$$

We observe that:

$$
(2k)^2 + 2k + \tfrac{1}{2} = (2k + 1)^2 - (2k + 1) + \tfrac{1}{2}
$$

$$
(2k)^2 - 2k + \tfrac{1}{2} = (2k - 1)^2 + (2k - 1) + \tfrac{1}{2}
$$

Thus, when the numerator and denominator are written as products of these terms, the factors in the numerator cancel with the corresponding factors in the denominator **except for the last two terms** in the numerator and the first term in the denominator.

---

### Step 3: Simplification

After cancellation, the resulting expression is:

$$
\frac{(4n^2 + 2n + \tfrac{1}{2})}{(\tfrac{1}{2})}
$$

Simplify the division:

$$
\frac{4n^2 + 2n + \tfrac{1}{2}}{\tfrac{1}{2}} = 2 \cdot (4n^2 + 2n + \tfrac{1}{2}) = 8n^2 + 4n + 1
$$

---

### Final Result

$$
\boxed{8n^2 + 4n + 1}
$$

[Assistant]
Explanation:
Okay, so I need to simplify this big fraction where the numerator is the product of terms like (2^4 + 1/4), (4^4 + 1/4), all the way up to (2n)^4 + 1/4, and the denominator is the product of similar terms but with odd numbers: (1^4 + 1/4), (3^4 + 1/4), ..., up to (2n - 1)^4 + 1/4.

First, let me write down the expression clearly to make sure I understand it:

$$
\frac{(2^4 + \frac{1}{4})(4^4 + \frac{1}{4}) \dotsm [(2n)^4 + \frac{1}{4}]}{(1^4 + \frac{1}{4})(3^4 + \frac{1}{4}) \dotsm [(2n - 1)^4 + \frac{1}{4}]}
$$

So, the numerator has terms for even numbers from 2 up to 2n, and the denominator has terms for odd numbers from 1 up to 2n - 1. Each term in both the numerator and denominator is of the form k^4 + 1/4, where k is an integer. My job is to simplify this entire expression.

Hmm, I remember that sometimes expressions like a^4 + 1/4 can be factored. Maybe I can factor each term in the numerator and denominator, and then see if there's some cancellation. Let me try that.

First, let me recall how to factor expressions like a^4 + b^4 or similar. Wait, here it's a^4 + 1/4. Let me think. Maybe I can write it as a^4 + (1/2)^2? But that's not a standard identity. Alternatively, perhaps using complex numbers? But that might complicate things. Alternatively, maybe I can use the Sophie Germain identity or something similar?

Wait, Sophie Germain identity is for a^4 + 4b^4, right? Which factors into (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2). But in our case, we have a^4 + (1/4). Let me check if this can be related to Sophie Germain's identity.

Let me try to manipulate the expression. Let me denote k^4 + 1/4. Let me write 1/4 as (1/2)^2. So, we have k^4 + (1/2)^2. Hmm, which is not directly a Sophie Germain identity. Maybe if I adjust it somehow?

Alternatively, maybe I can think of it as k^4 + 4*(1/4)^2? Wait, 4*(1/4)^2 is 4*(1/16) = 1/4. So, that's exactly the same. So, k^4 + 4*(1/4)^2. So then, using Sophie Germain identity, which is a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2).

So here, if we take a = k and b = 1/2, then:

k^4 + 4*(1/2)^4 = k^4 + 4*(1/16) = k^4 + 1/4. Wait, but 4*(1/2)^4 is 4*(1/16) = 1/4. So, indeed, the expression k^4 + 1/4 can be written as a^4 + 4b^4 where a = k and b = 1/2. Therefore, we can factor it as:

(k^2 + 2*(k)*(1/2) + 2*(1/2)^2)(k^2 - 2*(k)*(1/2) + 2*(1/2)^2)

Simplify each term:

First factor:

k^2 + 2*(k)*(1/2) + 2*(1/2)^2 = k^2 + k + 2*(1/4) = k^2 + k + 1/2

Second factor:

k^2 - 2*(k)*(1/2) + 2*(1/2)^2 = k^2 - k + 1/2

Therefore, we have:

k^4 + 1/4 = (k^2 + k + 1/2)(k^2 - k + 1/2)

So that's a useful factorization!

Therefore, each term in the numerator and denominator can be factored into two quadratic terms. Maybe this will allow for cancellation?

Let me write down that factorization:

For each term (k^4 + 1/4) = (k^2 + k + 1/2)(k^2 - k + 1/2).

So then, the entire numerator is the product over even k from 2 to 2n of (k^2 + k + 1/2)(k^2 - k + 1/2), and the denominator is the product over odd k from 1 to 2n - 1 of (k^2 + k + 1/2)(k^2 - k + 1/2). Therefore, if I write the entire fraction as:

[Product_{even k} (k^2 + k + 1/2)(k^2 - k + 1/2)] / [Product_{odd k} (k^2 + k + 1/2)(k^2 - k + 1/2)]

Which can be written as:

[Product_{even k} (k^2 + k + 1/2) * Product_{even k} (k^2 - k + 1/2)] / [Product_{odd k} (k^2 + k + 1/2) * Product_{odd k} (k^2 - k + 1/2)]

So maybe splitting the numerator and denominator into two separate products. Let me think about how to handle each of these. Let me consider the products of (k^2 + k + 1/2) and (k^2 - k + 1/2) separately for even and odd k.

Alternatively, perhaps there's some telescoping happening here. Let me try to see if the terms can be rearranged or related to each other.

First, let me note that for a general k, if I consider (k^2 + k + 1/2) and (k^2 - k + 1/2), these are similar to quadratic expressions. Maybe if we write them in terms of (k + something), or relate them to terms for other values of k?

Alternatively, let me note that for each term in the numerator and denominator, after factoring, they have these quadratic factors. Let me check for specific values of k. Maybe I can compute a few terms to see a pattern.

Let me take k as even numbers first. Let me pick k = 2:

For k=2: (2^2 + 2 + 1/2) = 4 + 2 + 0.5 = 6.5 = 13/2

(2^2 - 2 + 1/2) = 4 - 2 + 0.5 = 2.5 = 5/2

Similarly, for k=4:

(4^2 + 4 + 1/2) = 16 + 4 + 0.5 = 20.5 = 41/2

(4^2 - 4 + 1/2) = 16 - 4 + 0.5 = 12.5 = 25/2

Similarly, for odd k, say k=1:

(1^2 + 1 + 1/2) = 1 + 1 + 0.5 = 2.5 = 5/2

(1^2 - 1 + 1/2) = 1 - 1 + 0.5 = 0.5 = 1/2

For k=3:

(9 + 3 + 0.5) = 12.5 = 25/2

(9 - 3 + 0.5) = 6.5 = 13/2

Hmm, interesting. For k=2, (k^2 + k + 1/2) is 13/2 and (k^2 - k + 1/2) is 5/2. For k=1, which is odd, (k^2 + k + 1/2) is 5/2 and (k^2 - k + 1/2) is 1/2. Similarly, for k=3, odd, the first term is 25/2 and the second term is 13/2. Wait, for k=2 even, (k^2 + k + 1/2) is 13/2, and for k=3 odd, the (k^2 - k + 1/2) term is 13/2. So there's a connection here. Similarly, for k=4 even, the first term is 41/2, which is same as k=5 odd's (k^2 - k + 1/2) term? Let me check k=5:

(5^2 - 5 + 1/2) = 25 - 5 + 0.5 = 20.5 = 41/2. Yes! Exactly. Similarly, for k=1 odd, (k^2 - k + 1/2) = 1/2, which is same as for k=0? Wait, k=0? If I plug k=0 into (k^2 + k + 1/2) it would be 0 + 0 + 1/2 = 1/2. But k=0 is not in our original products. However, this suggests that the terms for even k in the numerator's first factor (k^2 + k + 1/2) are equal to the terms for odd k in the denominator's second factor (k^2 - k + 1/2). Similarly, for the other way around?

Wait, let me think. Let me consider the general case. Suppose I have an even number k = 2m. Then:

(k^2 + k + 1/2) = (4m² + 2m + 1/2)

Similarly, for the odd number k = 2m - 1:

(k^2 - k + 1/2) = [(2m - 1)^2 - (2m - 1) + 1/2]

Let me compute that:

(4m² - 4m + 1) - (2m - 1) + 1/2 = 4m² - 4m + 1 - 2m + 1 + 1/2 = 4m² - 6m + 2 + 1/2 = 4m² - 6m + 5/2

Wait, but how does that compare to the even term?

Wait, earlier for k=2 (m=1):

k^2 + k + 1/2 = 4 + 2 + 0.5 = 6.5 = 13/2. For the odd k=1 (m=1), (k^2 - k + 1/2) = 1 - 1 + 0.5 = 0.5 = 1/2. Wait, that doesn't seem to match. Wait earlier I thought there was a connection, but maybe my previous thought was incorrect. Wait, let me check again with k=2 (even) and k=3 (odd):

For k=2 (even): (k^2 + k + 1/2) = 4 + 2 + 0.5 = 6.5 = 13/2

For k=3 (odd): (k^2 - k + 1/2) = 9 - 3 + 0.5 = 6.5 = 13/2. Ah! So here, for k=2 (even) and k=3 (odd), the first factor of the even term is equal to the second factor of the odd term. Similarly, for k=4 (even), (k^2 + k + 1/2) = 16 + 4 + 0.5 = 20.5 = 41/2. For k=5 (odd), (k^2 - k + 1/2) = 25 - 5 + 0.5 = 20.5 = 41/2. So same here. Similarly, for k=1 (odd), (k^2 - k + 1/2) = 1 - 1 + 0.5 = 0.5 = 1/2. While for k=0 (even), (k^2 + k + 1/2) = 0 + 0 + 0.5 = 0.5 = 1/2. But k=0 isn't in our denominator.

So the idea is that for each even number k = 2m, the term (k^2 + k + 1/2) is equal to the term (k^2 - k + 1/2) for the next odd number, which is k = 2m + 1. Wait, but in the case of k=2 (even), which is m=1, the term is (k^2 + k + 1/2) = 13/2, which equals (k^2 - k + 1/2) for k=3 (odd). Similarly, for k=4 (even), m=2, the term (k^2 + k + 1/2) = 41/2 equals the term (k^2 - k + 1/2) for k=5 (odd).

Therefore, this seems to suggest that if we have the numerator terms (even k) with the first factor (k^2 + k + 1/2) and denominator terms (odd k) with the second factor (k^2 - k + 1/2), they might cancel out. Similarly, the second factor of the even k terms (k^2 - k + 1/2) would correspond to the first factor of the odd k terms?

Wait, let me check for k=2 (even) second factor: (k^2 - k + 1/2) = 4 - 2 + 0.5 = 2.5 = 5/2. Which is the same as for k=1 (odd) first factor (k^2 + k + 1/2) = 1 + 1 + 0.5 = 2.5 = 5/2. Similarly, for k=4 (even) second factor: 16 - 4 + 0.5 = 12.5 = 25/2, which is same as k=3 (odd) first factor: 9 + 3 + 0.5 = 12.5 = 25/2.

Therefore, this suggests that for each even term in the numerator, the first factor (k^2 + k + 1/2) is equal to the second factor of the next odd term in the denominator, and the second factor of the even term (k^2 - k + 1/2) is equal to the first factor of the previous odd term in the denominator.

So if we consider the entire products, there might be cancellation between numerator and denominator. Let me try to write this out more formally.

Let me denote:

For each even k = 2m (where m ranges from 1 to n), we have:

Term in numerator: (k^2 + k + 1/2)(k^2 - k + 1/2)

For each odd k = 2m - 1 (where m ranges from 1 to n), we have:

Term in denominator: (k^2 + k + 1/2)(k^2 - k + 1/2)

So the entire fraction is:

[Numerator: product_{m=1}^n ( ( (2m)^2 + 2m + 1/2 ) * ( (2m)^2 - 2m + 1/2 ) ) ] / [Denominator: product_{m=1}^n ( ( (2m - 1)^2 + (2m - 1) + 1/2 ) * ( (2m - 1)^2 - (2m - 1) + 1/2 ) ) ]

But as we saw, ( (2m)^2 + 2m + 1/2 ) equals ( (2m + 1)^2 - (2m + 1) + 1/2 ), since if we take k = 2m + 1 (which is odd), then k^2 - k + 1/2 = (2m + 1)^2 - (2m + 1) + 1/2 = 4m² + 4m + 1 - 2m - 1 + 1/2 = 4m² + 2m + 0 + 1/2 = 4m² + 2m + 1/2. Which is exactly equal to (2m)^2 + 2m + 1/2, since (2m)^2 is 4m².

Similarly, ( (2m)^2 - 2m + 1/2 ) = (2m - 1)^2 + (2m - 1) + 1/2. Let me check:

k = 2m - 1 (odd). Then k^2 + k + 1/2 = (2m - 1)^2 + (2m - 1) + 1/2 = 4m² - 4m + 1 + 2m - 1 + 1/2 = 4m² - 2m + 0 + 1/2 = 4m² - 2m + 1/2. Which is equal to (2m)^2 - 2m + 1/2.

Therefore, this means that for each even term in the numerator, the first factor (k^2 + k + 1/2) is equal to the second factor of the next odd term in the denominator. Similarly, the second factor of the even term (k^2 - k + 1/2) is equal to the first factor of the previous odd term in the denominator.

Therefore, when we take the entire product, these terms will cancel out with the denominator. Let me think about how the products would look.

Let me write the numerator's factors and denominator's factors:

Numerator has for each even k=2m:

Factor1: ( (2m)^2 + 2m + 1/2 )

Factor2: ( (2m)^2 - 2m + 1/2 )

Denominator has for each odd k=2m-1:

Factor1: ( (2m - 1)^2 + (2m - 1) + 1/2 )

Factor2: ( (2m - 1)^2 - (2m - 1) + 1/2 )

But as established, Factor1 of numerator (even) = Factor2 of denominator (next odd). Similarly, Factor2 of numerator (even) = Factor1 of denominator (previous odd).

Therefore, if we look at the entire product, when we take the numerator over the denominator, all these terms will telescope. Let's think about how.

Let me denote:

For the numerator:

Product over m=1 to n of [Factor1_even(m) * Factor2_even(m)]

For the denominator:

Product over m=1 to n of [Factor1_odd(m) * Factor2_odd(m)]

But since Factor1_even(m) = Factor2_odd(m+1), and Factor2_even(m) = Factor1_odd(m). Wait, let me check:

For Factor1_even(m) = (2m)^2 + 2m + 1/2. Which is equal to (2m + 1)^2 - (2m + 1) + 1/2. Which is the Factor2_odd(m+1).

Similarly, Factor2_even(m) = (2m)^2 - 2m + 1/2. Which is equal to (2m - 1)^2 + (2m - 1) + 1/2. Which is the Factor1_odd(m).

Therefore, if we consider the entire products:

Numerator: [Product_{m=1}^n (Factor1_even(m) * Factor2_even(m))]

Denominator: [Product_{m=1}^n (Factor1_odd(m) * Factor2_odd(m))]

But Factor1_even(m) = Factor2_odd(m+1) and Factor2_even(m) = Factor1_odd(m). Therefore, substituting into the numerator:

Product_{m=1}^n [Factor2_odd(m+1) * Factor1_odd(m)]

Which is equal to [Product_{m=1}^n Factor1_odd(m) * Product_{m=1}^n Factor2_odd(m+1)]

But Product_{m=1}^n Factor2_odd(m+1) = Product_{m=2}^{n+1} Factor2_odd(m)

Similarly, Product_{m=1}^n Factor1_odd(m) = Product_{m=1}^n Factor1_odd(m)

Therefore, the entire numerator becomes:

[Product_{m=1}^n Factor1_odd(m)] * [Product_{m=2}^{n+1} Factor2_odd(m)]

So when we divide by the denominator, which is [Product_{m=1}^n Factor1_odd(m) * Factor2_odd(m)], we get:

[Product_{m=1}^n Factor1_odd(m) * Product_{m=2}^{n+1} Factor2_odd(m)] / [Product_{m=1}^n Factor1_odd(m) * Product_{m=1}^n Factor2_odd(m)]

This simplifies to [Product_{m=2}^{n+1} Factor2_odd(m)] / [Product_{m=1}^n Factor2_odd(m)]

Which is equal to [Product_{m=2}^{n+1} Factor2_odd(m)] / [Product_{m=1}^n Factor2_odd(m)]

This simplifies further to [Factor2_odd(n+1) / Factor2_odd(1)]

Because when you divide the two products, the terms from m=2 to n cancel out, leaving Factor2_odd(n+1) divided by Factor2_odd(1).

So, the entire expression simplifies to Factor2_odd(n+1) divided by Factor2_odd(1).

So, what is Factor2_odd(m)? Recall that Factor2_odd(m) is (k^2 - k + 1/2) where k = 2m - 1.

Therefore, Factor2_odd(n+1) is [(2(n+1) - 1)^2 - (2(n+1) - 1) + 1/2] = [(2n + 1)^2 - (2n + 1) + 1/2]

Similarly, Factor2_odd(1) is [(2*1 - 1)^2 - (2*1 - 1) + 1/2] = [1^2 - 1 + 1/2] = 0 + 1/2 = 1/2

So let me compute Factor2_odd(n+1):

Compute [(2n + 1)^2 - (2n + 1) + 1/2]

First expand (2n + 1)^2 = 4n² + 4n + 1

Subtract (2n + 1): 4n² + 4n + 1 - 2n - 1 = 4n² + 2n

Add 1/2: 4n² + 2n + 1/2

So Factor2_odd(n+1) = 4n² + 2n + 1/2

Therefore, the entire expression simplifies to [4n² + 2n + 1/2] / [1/2]

Which is equal to [ (4n² + 2n + 1/2) ] / (1/2) = 2*(4n² + 2n + 1/2) = 8n² + 4n + 1

Wait, let me check that arithmetic again.

Wait, if you have (a) / (1/2) = 2a. So:

[4n² + 2n + 1/2] divided by 1/2 is equal to 2*(4n² + 2n + 1/2) = 8n² + 4n + 1.

Yes. So the entire expression simplifies to 8n² + 4n + 1.

But let me verify this with a small value of n to check if this is correct. Let's take n=1.

Case n=1:

Numerator: (2^4 + 1/4) = 16 + 1/4 = 65/4

Denominator: (1^4 + 1/4) = 1 + 1/4 = 5/4

So the entire expression is (65/4) / (5/4) = 65/5 = 13.

Now according to the formula 8n² + 4n + 1 with n=1: 8(1) + 4(1) + 1 = 8 + 4 + 1 = 13. Correct.

Another test case: n=2.

Numerator terms: (2^4 + 1/4)(4^4 + 1/4) = (16 + 0.25)(256 + 0.25) = (16.25)(256.25)

Denominator terms: (1^4 + 1/4)(3^4 + 1/4) = (1.25)(81 + 0.25) = 1.25 * 81.25

Compute numerator: 16.25 * 256.25

Let me compute that:

But maybe it's easier to use fractions.

Original terms:

For n=2:

Numerator: (2^4 + 1/4) = 16 + 1/4 = 65/4

(4^4 + 1/4) = 256 + 1/4 = 1025/4

So numerator product: (65/4) * (1025/4) = (65 * 1025)/16

Denominator: (1^4 + 1/4) = 5/4; (3^4 + 1/4) = 81 + 1/4 = 325/4. So denominator product: (5/4)*(325/4) = (5*325)/16

Therefore, the entire expression is [(65*1025)/16] / [(5*325)/16] = (65*1025)/(5*325)

Simplify:

Divide numerator and denominator by 5: (13*1025)/(325)

Note that 325 = 25*13, so:

(13*1025)/(25*13) = 1025/25 = 41.

Now according to the formula 8n² + 4n + 1 with n=2: 8*(4) + 4*2 + 1 = 32 + 8 + 1 = 41. Correct again!

Another check with n=3.

Compute formula: 8*(9) + 4*3 + 1 = 72 + 12 + 1 = 85.

Let me compute the original expression for n=3:

Numerator: (2^4 + 1/4)(4^4 + 1/4)(6^4 + 1/4) = (65/4)(1025/4)(1296 + 1/4) = (65/4)(1025/4)(5185/4)

Denominator: (1^4 + 1/4)(3^4 + 1/4)(5^4 + 1/4) = (5/4)(325/4)(625 + 1/4) = (5/4)(325/4)(2501/4)

Compute the entire expression:

[ (65/4)(1025/4)(5185/4) ] / [ (5/4)(325/4)(2501/4) ]

This simplifies to [65 * 1025 * 5185] / [5 * 325 * 2501]

Let me compute step by step:

First, 65 / 5 = 13

Then, 1025 / 325 = 1025 / 325. Let me divide numerator and denominator by 25: 41 / 13.

Then, 5185 / 2501. Let me check if 2501 * 2 = 5002, so 5185 - 5002 = 183, so maybe not. Alternatively, let me check 2501: 2501 is 50^2 + 1 = 2500 + 1 = 2501. Wait, and 5185 is 5185. Let me check if 2501 divides into 5185.

Wait, 2501 * 2 = 5002, 5185 - 5002 = 183. So no. Wait, but perhaps there's a relationship here. Let me recall that earlier, our telescoping led us to 8n² + 4n + 1. For n=3, it's 85. So the result should be 85. Let me check if [13 * (41/13) * (5185 / 2501)] equals 85.

Simplify 13 * (41/13) = 41. Then 41 * (5185 / 2501) = ?

Let me check 5185 divided by 2501. Let me compute 2501 * 2 = 5002. 5185 - 5002 = 183. So 5185 = 2501 * 2 + 183. Let me check if 183 relates to something. Wait, but if the result is supposed to be 85, then 41 * (5185 / 2501) = 85 implies 5185 / 2501 = 85 / 41. Let me compute 85 / 41 ≈ 2.073. Let me compute 5185 / 2501 ≈ 5185 / 2501 ≈ 2.073. Let me check 2501 * 85 = 2501 * 80 + 2501 * 5 = 200,080 + 12,505 = 212,585. And 41 * 5185 = 41 * 5000 + 41 * 185 = 205,000 + 7,585 = 212,585. Yes! Therefore, 41 * (5185 / 2501) = 85. Therefore, the total expression is indeed 85. Which matches the formula.

Therefore, the formula seems correct. Therefore, the simplified expression is 8n² + 4n + 1.

Wait, but let me check with n=1, 2, 3 and it works, so seems solid.

But let me also check if there is another way to write 8n² + 4n + 1. Maybe factoring?

8n² + 4n + 1. Let me see:

It doesn't factor nicely with integer coefficients. Let me check discriminant: 16 - 32 = -16? Wait no. Wait discriminant is b² - 4ac = 16 - 32 = -16? Wait no, 8n² + 4n + 1. Discriminant is 16 - 32 = -16. So it doesn't factor over reals? Wait, no, discriminant is 16 - 4*8*1 = 16 - 32 = -16. Therefore, complex roots. Therefore, the expression is irreducible over real numbers. Therefore, the simplified form is 8n² + 4n + 1.

But let me check again if the telescoping was correctly done. The key step was realizing that the numerator factors cancel with denominator factors, leaving only the last term in numerator and the first term in denominator? Wait, earlier we had that the entire expression simplified to Factor2_odd(n+1)/Factor2_odd(1). And Factor2_odd(n+1) was 4n² + 2n + 1/2, then divided by 1/2 gives 8n² + 4n + 1.

Yes. So that seems correct. Therefore, the simplified expression is 8n² + 4n + 1.

But let me check if there is another way to write this. For example, note that 8n² + 4n + 1 can be written as (2n + 1)^2 + (2n)^2? Let me check:

(2n + 1)^2 = 4n² + 4n + 1, and (2n)^2 = 4n². So sum is 8n² + 4n + 1. Yes. So 8n² + 4n + 1 = (2n)^2 + (2n + 1)^2. Interesting. But maybe that's not necessary.

Alternatively, maybe written as 4*(2n² + n) + 1, but again not sure. Anyway, the answer is 8n² + 4n + 1.

But let me check for n=1: 8 + 4 + 1 = 13, which matches. n=2: 32 + 8 + 1 = 41, which matches. n=3: 72 + 12 + 1 = 85. Correct.

Alternatively, maybe there's a different expression? Let me think again.

Wait, let me think back to the original factors.

We had the entire expression simplified to [Factor2_odd(n+1)] / [Factor2_odd(1)].

Factor2_odd(n+1) was [(2n + 1)^2 - (2n + 1) + 1/2] = 4n² + 2n + 1/2.

But if I compute 4n² + 2n + 1/2 divided by 1/2, it's 8n² + 4n + 1.

Alternatively, maybe there's a different way to express this? For example, if I consider that the entire original expression was telescoped to a single term, perhaps related to (some term with n)?

But given that the test cases work, and the derivation seems correct, I think that 8n² + 4n + 1 is the correct simplified expression. Therefore, the answer is \boxed{8n^2 + 4n + 1}.

But wait, let me check once more for n=1,2,3. For n=1: 8(1)^2 + 4(1) + 1 = 13, which matched. For n=2: 8*4 + 8 +1 = 32 + 8 + 1 = 41, correct. For n=3: 8*9 + 12 + 1 = 72 + 12 + 1 = 85. Correct.

Alternatively, is there a possibility that this expression can be simplified more? Let me check if 8n² + 4n + 1 can be written as (something else). For example, maybe related to (2n + 1)^2 + something?

(2n + 1)^2 = 4n² + 4n + 1. So 8n² + 4n + 1 = 2*(4n² + 2n) + 1. Hmm, not sure. Alternatively, maybe 8n² + 4n + 1 = 4n² + (4n² + 4n + 1) = 4n² + (2n + 1)^2. Which is another way to write it. But I think the problem just wants the simplified expression, and 8n² + 4n + 1 is a valid simplified form.

Therefore, I think that the answer is 8n² + 4n + 1.

**Final Answer**
\boxed{8n^2 + 4n + 1}

To simplify the given expression:

$$
\frac{(2^4 + \frac{1}{4})(4^4 + \frac{1}{4}) \dotsm [(2n)^4 + \frac{1}{4}]}{(1^4 + \frac{1}{4})(3^4 + \frac{1}{4}) \dotsm [(2n - 1)^4 + \frac{1}{4}]}
$$

we observe that each term in the numerator is of the form $(k^4 + \tfrac{1}{4})$ where $k = 2, 4, 6, \ldots, 2n$, and each term in the denominator is of the form $(k^4 + \tfrac{1}{4})$ where $k = 1, 3, 5, \ldots, 2n - 1$.

---

### Step 1: Factorization using Sophie Germain Identity

We use the identity:

$$
a^4 + 4b^4 = (a^2 + 2ab + 2b^2)(a^2 - 2ab + 2b^2)
$$

Let us apply this identity to $k^4 + \tfrac{1}{4}$ by writing it as:

$$
k^4 + \frac{1}{4} = k^4 + 4\left(\frac{1}{4}\right)^2
$$

So let $a = k$, $b = \frac{1}{2}$, then:

$$
k^4 + \frac{1}{4} = (k^2 + 2k \cdot \frac{1}{2} + 2 \cdot \left(\frac{1}{2}\right)^2)(k^2 - 2k \cdot \frac{1}{2} + 2 \cdot \left(\frac{1}{2}\right)^2)
$$

Simplifying:

$$
k^4 + \frac{1}{4} = (k^2 + k + \tfrac{1}{2})(k^2 - k + \tfrac{1}{2})
$$

Thus, every factor in both the numerator and denominator can be written as a product of these two quadratic expressions.

---

### Step 2: Telescoping Structure

Now write the entire numerator and denominator using the factorization:

- **Numerator:**

$$
\prod_{k=1}^n ( (2k)^2 + 2k + \tfrac{1}{2} ) \cdot ( (2k)^2 - 2k + \tfrac{1}{2} )
$$

- **Denominator:**

$$
\prod_{k=1}^n ( (2k - 1)^2 + (2k - 1) + \tfrac{1}{2} ) \cdot ( (2k - 1)^2 - (2k - 1) + \tfrac{1}{2} )
$$

We observe that:

$$
(2k)^2 + 2k + \tfrac{1}{2} = (2k + 1)^2 - (2k + 1) + \tfrac{1}{2}
$$

$$
(2k)^2 - 2k + \tfrac{1}{2} = (2k - 1)^2 + (2k - 1) + \tfrac{1}{2}
$$

Thus, when the numerator and denominator are written as products of these terms, the factors in the numerator cancel with the corresponding factors in the denominator **except for the last two terms** in the numerator and the first term in the denominator.

---

### Step 3: Simplification

After cancellation, the resulting expression is:

$$
\frac{(4n^2 + 2n + \tfrac{1}{2})}{(\tfrac{1}{2})}
$$

Simplify the division:

$$
\frac{4n^2 + 2n + \tfrac{1}{2}}{\tfrac{1}{2}} = 2 \cdot (4n^2 + 2n + \tfrac{1}{2}) = 8n^2 + 4n + 1
$$

---

### Final Result

$$
\boxed{8n^2 + 4n + 1}
$$

Solution:
Let
\[f(m) = m^4 + \frac{1}{4} = \frac{4m^4 + 1}{4}.\]We can factor this with a little give and take:
\begin{align*}
f(m) &= \frac{4m^4 + 1}{4} \\
&= \frac{4m^4 + 4m^2 + 1 - 4m^2}{4} \\
&= \frac{(2m^2 + 1)^2 - (2m)^2}{4} \\
&= \frac{(2m^2 + 2m + 1)(2m^2 - 2m + 1)}{4}.
\end{align*}Now, let $g(m) = 2m^2 + 2m + 1.$  Then
\[g(m - 1) = 2(m - 1)^2 + 2(m - 1) + 1 = 2m^2 - 2m + 1.\]Hence,
\[f(m) = \frac{g(m) g(m - 1)}{4}.\]Therefore,
\begin{align*}
\frac{(2^4 + \frac{1}{4})(4^4 + \frac{1}{4}) \dotsm [(2n)^4 + \frac{1}{4}]}{(1^4 + \frac{1}{4})(3^4 + \frac{1}{4}) \dotsm [(2n - 1)^4 + \frac{1}{4}]} &= \frac{f(2) f(4) \dotsm f(2n)}{f(1) f(3) \dotsm f(2n - 1)} \\
&= \frac{\frac{g(2) g(1)}{4} \cdot \frac{g(4) g(3)}{4} \dotsm \frac{g(2n) g(2n - 1)}{4}}{\frac{g(1) g(0)}{4} \cdot \frac{g(3) g(2)}{4} \dotsm \frac{g(2n - 1) g(2n - 2)}{4}} \\
&= \frac{g(2n)}{g(0)} \\
&= 2(2n)^2 + 2(2n) + 1 \\
&= \boxed{8n^2 + 4n + 1}.
\end{align*}


Explanation:
Okay, so I need to find the probability that the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal. Then express that probability as a reduced fraction a/b and find a + b. Alright, let's start by recalling what makes a decimal terminating.

From what I remember, a decimal expansion of a fraction terminates if and only if the denominator, after simplifying the fraction to its lowest terms, has no prime factors other than 2 and 5. So, for 1/n to have a terminating decimal, n must be of the form 2^k * 5^m where k and m are non-negative integers.

But wait, in this problem, we are dealing with reciprocals of odd integers. So the numbers we are considering are odd, which means they are not divisible by 2. Therefore, for the reciprocal 1/n to have a terminating decimal, the denominator n must only have prime factors 2 and 5. However, since n is odd, it cannot have 2 as a prime factor. Therefore, n must be a power of 5. Because if it's only allowed to have 2 and 5 as prime factors, but since it's odd, it can't have 2, so it must be a power of 5. So, the numbers n are the odd integers less than 2010 which are powers of 5.

Wait, is that correct? Let me think again. Suppose n is an odd integer, so n is not divisible by 2. For 1/n to be terminating, n must be of the form 2^k * 5^m, but since n is odd, k must be zero, so n must be 5^m. Therefore, yes, n must be a power of 5. Therefore, the numbers we need are the odd integers less than 2010 that are powers of 5. Therefore, the problem reduces to counting how many such numbers there are, then dividing by the total number of positive odd integers less than 2010. Then we have to simplify that fraction and find a + b.

So first, let's find the total number of positive odd integers less than 2010. Well, the positive odd integers less than 2010 are 1, 3, 5, ..., up to the largest odd number less than 2010. Since 2010 is even, the largest odd integer less than 2010 is 2009. The number of terms in this sequence is equal to (2009 - 1)/2 + 1 = (2008)/2 + 1 = 1004 + 1 = 1005. Wait, let me check that formula again. The formula for the number of terms in an arithmetic sequence is ((last term - first term)/common difference) + 1. So here, first term is 1, last term is 2009, common difference is 2. So ((2009 - 1)/2) + 1 = (2008 / 2) + 1 = 1004 + 1 = 1005. Yes, so there are 1005 positive odd integers less than 2010. So the denominator of the probability will be 1005.

Now, the numerator is the number of positive odd integers less than 2010 that are powers of 5. So we need to find all numbers of the form 5^k where 5^k < 2010 and 5^k is odd. Since 5 is odd, any power of 5 is odd, so we don't have to worry about that. So we need to find k such that 5^k < 2010. Let me compute the powers of 5:

5^1 = 5

5^2 = 25

5^3 = 125

5^4 = 625

5^5 = 3125

Wait, 5^5 is 3125, which is greater than 2010, so the exponents go up to 4. Therefore, there are 4 such numbers: 5^1, 5^2, 5^3, 5^4, which are 5, 25, 125, 625. Wait, but 5^4 is 625, which is less than 2010. Then 5^5 is 3125, which is more than 2010, so that's excluded. Therefore, there are 4 numbers. Wait, but hold on. Wait 5^0 is 1, which is also a power of 5, right? Since 5^0 = 1. But the problem says "positive odd integers less than 2010". So 1 is included. So 5^0 is 1. So does that count?

Wait, the question says "positive odd integer less than 2010". So 1 is a positive odd integer less than 2010, so if we consider n=1, then 1/1 is 1.0, which is a terminating decimal. So yes, 1 should be included. So then, 5^0 is 1, so we have to count that as well. Therefore, the exponents k start from 0. Let me check 5^0 = 1, 5^1=5, 5^2=25, 5^3=125, 5^4=625, 5^5=3125. So 5^5 is 3125 which is greater than 2010. So k can be 0,1,2,3,4. Therefore, there are 5 numbers. Wait, so earlier I thought 4, but including 1, it's 5 numbers. So then the numerator is 5?

Wait, let me check if 1 is considered. The reciprocal of 1 is 1.0, which is terminating. So yes, 1 is indeed a valid number. Therefore, there are 5 such numbers: 1, 5, 25, 125, 625. Therefore, the numerator is 5, and the denominator is 1005, so the probability is 5/1005. Then we have to simplify this fraction.

But wait, let me check again. Maybe I missed something. Let me verify the powers of 5 again:

5^0 = 1

5^1 = 5

5^2 = 25

5^3 = 125

5^4 = 625

5^5 = 3125

Yes, 3125 is more than 2010, so up to 5^4. Therefore, exponents 0 through 4, inclusive. So that is 5 numbers. Therefore, 5 numbers. Therefore, probability is 5/1005.

But wait, 5 and 1005. Let me check if they have common factors. Let's compute GCD(5,1005). Well, 1005 divided by 5 is 201, so 1005 = 5*201. Therefore, 5/1005 reduces to 1/201. Therefore, the fraction is 1/201, so a=1, b=201, so a + b is 202. Therefore, the answer is 202?

Wait, but hold on. Let me think again. Is there any other number besides powers of 5 that could result in a terminating decimal when taking reciprocal?

Wait, earlier logic was that if n is odd, then n must be of the form 2^k * 5^m. But since it's odd, 2^k must be 1, so k=0, so n must be 5^m. Therefore, only powers of 5. Therefore, only numbers 1, 5, 25, 125, 625. So that's 5 numbers. Therefore, 5/1005 = 1/201. So a + b is 1 + 201 = 202. So is that the answer? Wait, but maybe there is something wrong here?

Wait, let me think again. Suppose n is an odd integer, so n is not divisible by 2. For 1/n to have a terminating decimal, n must have only 2 and 5 as prime factors. But since n is odd, it cannot have 2 as a prime factor. Therefore, n must be a power of 5. So that is correct. Therefore, the numbers are powers of 5. But let me check if there are other numbers. For example, suppose n is 1, which is 5^0, so that is included. Then 5, 25, 125, 625. So 5 numbers. Therefore, 5 numbers. So probability is 5 over total numbers of odd integers less than 2010, which is 1005. So 5/1005 reduces to 1/201, which is already reduced, so a=1, b=201, so a + b is 202. Therefore, is the answer 202?

But wait, I feel like maybe I'm missing something. Let me check with an example. Let me take a smaller number. Let's say instead of less than 2010, let's say less than 10. Then the odd numbers less than 10 are 1, 3, 5, 7, 9. Then the numbers which are powers of 5 are 1, 5. Therefore, probability is 2/5. Now, check if reciprocals of 1, 5 are terminating decimals. 1/1 = 1.0, terminating. 1/5 = 0.2, terminating. 1/3 is 0.333..., non-terminating. 1/7 is 0.142857..., non-terminating. 1/9 is 0.111..., non-terminating. So yes, only 1 and 5. Therefore, 2/5. So if we applied our formula, total numbers would be 5, and numbers which are powers of 5: 1, 5, so 2 numbers. Then 2/5, which reduces to 2/5. If we follow the same logic as above, for numbers less than 10, the answer would be 2/5. So that seems correct. So in that case, the logic holds.

Similarly, if we take numbers less than 25. Odd numbers less than 25: 1,3,5,...,23. Total of 12 numbers. Then numbers which are powers of 5: 1,5,25. But 25 is not less than 25, so only 1 and 5. Therefore, 2 numbers. Wait, but 25 is equal to 25, which is not less than 25. So, for numbers less than 25, the numbers would be 1,5. But in our case, the original problem says less than 2010. So, in that case, if the upper bound is 25, the numbers would be 1,5. But if upper bound is 25, then 25 is included? Wait, the problem says "less than 2010", so 2010 is excluded. So if we have numbers less than N, then numbers up to N - 1.

Therefore, in our original problem, numbers less than 2010, so up to 2009. So 5^4 = 625 is less than 2010. 5^5 = 3125 is greater than 2010, so excluded. So 1,5,25,125,625. So 5 numbers. Therefore, that seems correct.

But let me check if there's another way to think about this. Maybe numbers where the denominator after simplifying is 5^k? But since we are taking reciprocal of n, which is already in the form of 1/n. So the decimal terminates if and only if n is of the form 2^k * 5^m, but since n is odd, 2^k must be 1, so n is 5^m. Therefore, that is correct.

Wait, but what about numbers that are multiples of 5? For example, 15. 1/15. Let me check if 1/15 is terminating. 1 divided by 15 is 0.0666..., which is repeating. Therefore, not terminating. So even though 15 is a multiple of 5, since it's also a multiple of 3, which is not 2 or 5, so it doesn't work. Therefore, only numbers that are powers of 5. Hence, the conclusion is correct.

Therefore, the answer is 5/1005 reduces to 1/201, so a + b = 1 + 201 = 202. Therefore, the answer is 202. Wait, but before I conclude, let me check once again if there are more numbers.

Wait, is 1 considered a power of 5? Since 5^0 is 1. So if the problem says "positive odd integers", then 1 is included. So 1 is allowed. So the numbers are 5^0, 5^1, 5^2, 5^3, 5^4. Therefore, 5 numbers. Therefore, 5 numbers. Therefore, 5/1005 reduces to 1/201.

Wait, another way to check: total odd numbers less than 2010 is 1005. Then the number of favorable is 5. So 5/1005. Simplify numerator and denominator by 5: 1/201. So yes, that seems correct.

But maybe there is a mistake in counting the number of odd integers less than 2010? Let me confirm that again. The number of positive odd integers less than N is floor((N-1)/2) + 1? Wait, let me think. For example, if N is even, like 10. The number of odd numbers less than 10 is 5: 1,3,5,7,9. Which is (10 - 1)/2 = 4.5, floor(4.5) is 4, plus 1 is 5. So yes. So for N=2010, which is even, the number of odd numbers less than 2010 is floor((2010 - 1)/2) + 1? Wait, actually, the formula is ceiling(N/2) if N is even? Wait, let me think again.

Number of odd numbers less than N is equal to the number of odd numbers from 1 to N-1. Since N is even, N-1 is odd. The number of odd numbers from 1 to M is (M + 1)/2 if M is odd. Since N-1 is odd, so M = N - 1. Therefore, number is (M + 1)/2 = (N - 1 + 1)/2 = N/2. Therefore, for N even, the number of odd numbers less than N is N/2. Therefore, for N=2010, it's 2010 / 2 = 1005. Therefore, that's correct. Therefore, total numbers is 1005, which matches our previous calculation. Therefore, that's solid.

So then, 5 numbers in numerator. Therefore, 5/1005. Simplify: divide numerator and denominator by 5. 5 divided by 5 is 1. 1005 divided by 5 is 201. Therefore, 1/201. So a=1, b=201. Therefore, a + b is 202. Therefore, the answer is 202. Therefore, I think that's correct. But let me check once again for any possible mistake.

Wait, maybe I missed some other numbers? For example, numbers like 1, which is 5^0. What if there are other numbers with denominators that can be simplified? Wait, but the reciprocal is 1/n. So if n is already in its simplest form (which it is, since n is an integer). Therefore, if n has any prime factors other than 5, then 1/n won't terminate. Therefore, only numbers n which are powers of 5. Therefore, only 1,5,25,125,625. So that's 5 numbers.

Alternatively, maybe if n is 1? Well, 1 is allowed, and 1 is 5^0, so it's included. So that's correct. Therefore, the numerator is 5. Therefore, I think that's correct. Therefore, 5/1005 reduces to 1/201, so answer is 202.

But just to make sure, maybe there's another way to think about it. For example, suppose there was a number like 1/ (some multiple of 5 and 2s), but since the number is odd, you can't have 2s. So 2s are excluded. Therefore, only 5s. Therefore, yes, only powers of 5. Therefore, the answer is correct.

Therefore, I think that the answer is 202.

**Final Answer**
\boxed{202}

To solve this problem, we want to determine the probability that the reciprocal of a randomly selected **positive odd integer less than 2010** results in a **terminating decimal**. Once we find this probability in the form of a reduced fraction $ \frac{a}{b} $, we are to compute $ a + b $.

---

### 🔢 Step 1: Understanding Terminating Decimals

A rational number $ \frac{1}{n} $ has a terminating decimal representation **if and only if** the prime factorization of $ n $ (in its lowest form) contains **only the primes 2 and 5**.

Since we are examining the **reciprocal of an odd integer**, the number $ n $ must be **odd**, and so it **cannot** include the prime factor 2. Therefore, for $ \frac{1}{n} $ to be a terminating decimal, $ n $ must be a power of **5** — because that is the **only other allowed prime factor** in such a case.

Thus, we are looking for all **odd integers less than 2010** that are **powers of 5**.

---

### 🔢 Step 2: Count the Total Number of Odd Integers Less than 2010

Odd integers from 1 to 2009 form an arithmetic sequence:
- First term = 1
- Last term = 2009
- Common difference = 2

The number of terms in this sequence is:
$$
\text{Count} = \frac{2009 - 1}{2} + 1 = \frac{2008}{2} + 1 = 1004 + 1 = 1005
$$

So, there are **1005** positive odd integers less than 2010.

---

### 🔢 Step 3: Count the Favorable Numbers

We now look for **positive powers of 5** less than 2010.

Compute successive powers of 5:
- $ 5^0 = 1 $
- $ 5^1 = 5 $
- $ 5^2 = 25 $
- $ 5^3 = 125 $
- $ 5^4 = 625 $
- $ 5^5 = 3125 $ (too large, since 3125 > 2010)

So, the favorable numbers are:
$$
\{1, 5, 25, 125, 625\}
$$

That’s a total of **5** favorable numbers.

---

### 🔢 Step 4: Compute the Probability

We calculate the probability as:
$$
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{1005}
$$

Simplify the fraction:
- GCD of 5 and 1005 is 5
- $ \frac{5}{1005} = \frac{1}{201} $

So, the fraction in lowest terms is:
$$
\frac{a}{b} = \frac{1}{201}
$$

---

### 🔢 Step 5: Compute $ a + b $

From the reduced fraction, we have:
- $ a = 1 $
- $ b = 201 $

Therefore:
$$
a + b = 1 + 201 = \boxed{202}
$$

[Assistant]
Explanation:
Okay, so I need to find the probability that the reciprocal of a randomly selected positive odd integer less than 2010 gives a terminating decimal. Then express that probability as a reduced fraction a/b and find a + b. Alright, let's start by recalling what makes a decimal terminating.

From what I remember, a decimal expansion of a fraction terminates if and only if the denominator, after simplifying the fraction to its lowest terms, has no prime factors other than 2 and 5. So, for 1/n to have a terminating decimal, n must be of the form 2^k * 5^m where k and m are non-negative integers.

But wait, in this problem, we are dealing with reciprocals of odd integers. So the numbers we are considering are odd, which means they are not divisible by 2. Therefore, for the reciprocal 1/n to have a terminating decimal, the denominator n must only have prime factors 2 and 5. However, since n is odd, it cannot have 2 as a prime factor. Therefore, n must be a power of 5. Because if it's only allowed to have 2 and 5 as prime factors, but since it's odd, it can't have 2, so it must be a power of 5. So, the numbers n are the odd integers less than 2010 which are powers of 5.

Wait, is that correct? Let me think again. Suppose n is an odd integer, so n is not divisible by 2. For 1/n to be terminating, n must be of the form 2^k * 5^m, but since n is odd, k must be zero, so n must be 5^m. Therefore, yes, n must be a power of 5. Therefore, the numbers we need are the odd integers less than 2010 that are powers of 5. Therefore, the problem reduces to counting how many such numbers there are, then dividing by the total number of positive odd integers less than 2010. Then we have to simplify that fraction and find a + b.

So first, let's find the total number of positive odd integers less than 2010. Well, the positive odd integers less than 2010 are 1, 3, 5, ..., up to the largest odd number less than 2010. Since 2010 is even, the largest odd integer less than 2010 is 2009. The number of terms in this sequence is equal to (2009 - 1)/2 + 1 = (2008)/2 + 1 = 1004 + 1 = 1005. Wait, let me check that formula again. The formula for the number of terms in an arithmetic sequence is ((last term - first term)/common difference) + 1. So here, first term is 1, last term is 2009, common difference is 2. So ((2009 - 1)/2) + 1 = (2008 / 2) + 1 = 1004 + 1 = 1005. Yes, so there are 1005 positive odd integers less than 2010. So the denominator of the probability will be 1005.

Now, the numerator is the number of positive odd integers less than 2010 that are powers of 5. So we need to find all numbers of the form 5^k where 5^k < 2010 and 5^k is odd. Since 5 is odd, any power of 5 is odd, so we don't have to worry about that. So we need to find k such that 5^k < 2010. Let me compute the powers of 5:

5^1 = 5

5^2 = 25

5^3 = 125

5^4 = 625

5^5 = 3125

Wait, 5^5 is 3125, which is greater than 2010, so the exponents go up to 4. Therefore, there are 4 such numbers: 5^1, 5^2, 5^3, 5^4, which are 5, 25, 125, 625. Wait, but 5^4 is 625, which is less than 2010. Then 5^5 is 3125, which is more than 2010, so that's excluded. Therefore, there are 4 numbers. Wait, but hold on. Wait 5^0 is 1, which is also a power of 5, right? Since 5^0 = 1. But the problem says "positive odd integers less than 2010". So 1 is included. So 5^0 is 1. So does that count?

Wait, the question says "positive odd integer less than 2010". So 1 is a positive odd integer less than 2010, so if we consider n=1, then 1/1 is 1.0, which is a terminating decimal. So yes, 1 should be included. So then, 5^0 is 1, so we have to count that as well. Therefore, the exponents k start from 0. Let me check 5^0 = 1, 5^1=5, 5^2=25, 5^3=125, 5^4=625, 5^5=3125. So 5^5 is 3125 which is greater than 2010. So k can be 0,1,2,3,4. Therefore, there are 5 numbers. Wait, so earlier I thought 4, but including 1, it's 5 numbers. So then the numerator is 5?

Wait, let me check if 1 is considered. The reciprocal of 1 is 1.0, which is terminating. So yes, 1 is indeed a valid number. Therefore, there are 5 such numbers: 1, 5, 25, 125, 625. Therefore, the numerator is 5, and the denominator is 1005, so the probability is 5/1005. Then we have to simplify this fraction.

But wait, let me check again. Maybe I missed something. Let me verify the powers of 5 again:

5^0 = 1

5^1 = 5

5^2 = 25

5^3 = 125

5^4 = 625

5^5 = 3125

Yes, 3125 is more than 2010, so up to 5^4. Therefore, exponents 0 through 4, inclusive. So that is 5 numbers. Therefore, 5 numbers. Therefore, probability is 5/1005.

But wait, 5 and 1005. Let me check if they have common factors. Let's compute GCD(5,1005). Well, 1005 divided by 5 is 201, so 1005 = 5*201. Therefore, 5/1005 reduces to 1/201. Therefore, the fraction is 1/201, so a=1, b=201, so a + b is 202. Therefore, the answer is 202?

Wait, but hold on. Let me think again. Is there any other number besides powers of 5 that could result in a terminating decimal when taking reciprocal?

Wait, earlier logic was that if n is odd, then n must be of the form 2^k * 5^m. But since it's odd, 2^k must be 1, so k=0, so n must be 5^m. Therefore, only powers of 5. Therefore, only numbers 1, 5, 25, 125, 625. So that's 5 numbers. Therefore, 5/1005 = 1/201. So a + b is 1 + 201 = 202. So is that the answer? Wait, but maybe there is something wrong here?

Wait, let me think again. Suppose n is an odd integer, so n is not divisible by 2. For 1/n to have a terminating decimal, n must have only 2 and 5 as prime factors. But since n is odd, it cannot have 2 as a prime factor. Therefore, n must be a power of 5. So that is correct. Therefore, the numbers are powers of 5. But let me check if there are other numbers. For example, suppose n is 1, which is 5^0, so that is included. Then 5, 25, 125, 625. So 5 numbers. Therefore, 5 numbers. So probability is 5 over total numbers of odd integers less than 2010, which is 1005. So 5/1005 reduces to 1/201, which is already reduced, so a=1, b=201, so a + b is 202. Therefore, is the answer 202?

But wait, I feel like maybe I'm missing something. Let me check with an example. Let me take a smaller number. Let's say instead of less than 2010, let's say less than 10. Then the odd numbers less than 10 are 1, 3, 5, 7, 9. Then the numbers which are powers of 5 are 1, 5. Therefore, probability is 2/5. Now, check if reciprocals of 1, 5 are terminating decimals. 1/1 = 1.0, terminating. 1/5 = 0.2, terminating. 1/3 is 0.333..., non-terminating. 1/7 is 0.142857..., non-terminating. 1/9 is 0.111..., non-terminating. So yes, only 1 and 5. Therefore, 2/5. So if we applied our formula, total numbers would be 5, and numbers which are powers of 5: 1, 5, so 2 numbers. Then 2/5, which reduces to 2/5. If we follow the same logic as above, for numbers less than 10, the answer would be 2/5. So that seems correct. So in that case, the logic holds.

Similarly, if we take numbers less than 25. Odd numbers less than 25: 1,3,5,...,23. Total of 12 numbers. Then numbers which are powers of 5: 1,5,25. But 25 is not less than 25, so only 1 and 5. Therefore, 2 numbers. Wait, but 25 is equal to 25, which is not less than 25. So, for numbers less than 25, the numbers would be 1,5. But in our case, the original problem says less than 2010. So, in that case, if the upper bound is 25, the numbers would be 1,5. But if upper bound is 25, then 25 is included? Wait, the problem says "less than 2010", so 2010 is excluded. So if we have numbers less than N, then numbers up to N - 1.

Therefore, in our original problem, numbers less than 2010, so up to 2009. So 5^4 = 625 is less than 2010. 5^5 = 3125 is greater than 2010, so excluded. So 1,5,25,125,625. So 5 numbers. Therefore, that seems correct.

But let me check if there's another way to think about this. Maybe numbers where the denominator after simplifying is 5^k? But since we are taking reciprocal of n, which is already in the form of 1/n. So the decimal terminates if and only if n is of the form 2^k * 5^m, but since n is odd, 2^k must be 1, so n is 5^m. Therefore, that is correct.

Wait, but what about numbers that are multiples of 5? For example, 15. 1/15. Let me check if 1/15 is terminating. 1 divided by 15 is 0.0666..., which is repeating. Therefore, not terminating. So even though 15 is a multiple of 5, since it's also a multiple of 3, which is not 2 or 5, so it doesn't work. Therefore, only numbers that are powers of 5. Hence, the conclusion is correct.

Therefore, the answer is 5/1005 reduces to 1/201, so a + b = 1 + 201 = 202. Therefore, the answer is 202. Wait, but before I conclude, let me check once again if there are more numbers.

Wait, is 1 considered a power of 5? Since 5^0 is 1. So if the problem says "positive odd integers", then 1 is included. So 1 is allowed. So the numbers are 5^0, 5^1, 5^2, 5^3, 5^4. Therefore, 5 numbers. Therefore, 5 numbers. Therefore, 5/1005 reduces to 1/201.

Wait, another way to check: total odd numbers less than 2010 is 1005. Then the number of favorable is 5. So 5/1005. Simplify numerator and denominator by 5: 1/201. So yes, that seems correct.

But maybe there is a mistake in counting the number of odd integers less than 2010? Let me confirm that again. The number of positive odd integers less than N is floor((N-1)/2) + 1? Wait, let me think. For example, if N is even, like 10. The number of odd numbers less than 10 is 5: 1,3,5,7,9. Which is (10 - 1)/2 = 4.5, floor(4.5) is 4, plus 1 is 5. So yes. So for N=2010, which is even, the number of odd numbers less than 2010 is floor((2010 - 1)/2) + 1? Wait, actually, the formula is ceiling(N/2) if N is even? Wait, let me think again.

Number of odd numbers less than N is equal to the number of odd numbers from 1 to N-1. Since N is even, N-1 is odd. The number of odd numbers from 1 to M is (M + 1)/2 if M is odd. Since N-1 is odd, so M = N - 1. Therefore, number is (M + 1)/2 = (N - 1 + 1)/2 = N/2. Therefore, for N even, the number of odd numbers less than N is N/2. Therefore, for N=2010, it's 2010 / 2 = 1005. Therefore, that's correct. Therefore, total numbers is 1005, which matches our previous calculation. Therefore, that's solid.

So then, 5 numbers in numerator. Therefore, 5/1005. Simplify: divide numerator and denominator by 5. 5 divided by 5 is 1. 1005 divided by 5 is 201. Therefore, 1/201. So a=1, b=201. Therefore, a + b is 202. Therefore, the answer is 202. Therefore, I think that's correct. But let me check once again for any possible mistake.

Wait, maybe I missed some other numbers? For example, numbers like 1, which is 5^0. What if there are other numbers with denominators that can be simplified? Wait, but the reciprocal is 1/n. So if n is already in its simplest form (which it is, since n is an integer). Therefore, if n has any prime factors other than 5, then 1/n won't terminate. Therefore, only numbers n which are powers of 5. Therefore, only 1,5,25,125,625. So that's 5 numbers.

Alternatively, maybe if n is 1? Well, 1 is allowed, and 1 is 5^0, so it's included. So that's correct. Therefore, the numerator is 5. Therefore, I think that's correct. Therefore, 5/1005 reduces to 1/201, so answer is 202.

But just to make sure, maybe there's another way to think about it. For example, suppose there was a number like 1/ (some multiple of 5 and 2s), but since the number is odd, you can't have 2s. So 2s are excluded. Therefore, only 5s. Therefore, yes, only powers of 5. Therefore, the answer is correct.

Therefore, I think that the answer is 202.

**Final Answer**
\boxed{202}

To solve this problem, we want to determine the probability that the reciprocal of a randomly selected **positive odd integer less than 2010** results in a **terminating decimal**. Once we find this probability in the form of a reduced fraction $ \frac{a}{b} $, we are to compute $ a + b $.

---

### 🔢 Step 1: Understanding Terminating Decimals

A rational number $ \frac{1}{n} $ has a terminating decimal representation **if and only if** the prime factorization of $ n $ (in its lowest form) contains **only the primes 2 and 5**.

Since we are examining the **reciprocal of an odd integer**, the number $ n $ must be **odd**, and so it **cannot** include the prime factor 2. Therefore, for $ \frac{1}{n} $ to be a terminating decimal, $ n $ must be a power of **5** — because that is the **only other allowed prime factor** in such a case.

Thus, we are looking for all **odd integers less than 2010** that are **powers of 5**.

---

### 🔢 Step 2: Count the Total Number of Odd Integers Less than 2010

Odd integers from 1 to 2009 form an arithmetic sequence:
- First term = 1
- Last term = 2009
- Common difference = 2

The number of terms in this sequence is:
$$
\text{Count} = \frac{2009 - 1}{2} + 1 = \frac{2008}{2} + 1 = 1004 + 1 = 1005
$$

So, there are **1005** positive odd integers less than 2010.

---

### 🔢 Step 3: Count the Favorable Numbers

We now look for **positive powers of 5** less than 2010.

Compute successive powers of 5:
- $ 5^0 = 1 $
- $ 5^1 = 5 $
- $ 5^2 = 25 $
- $ 5^3 = 125 $
- $ 5^4 = 625 $
- $ 5^5 = 3125 $ (too large, since 3125 > 2010)

So, the favorable numbers are:
$$
\{1, 5, 25, 125, 625\}
$$

That’s a total of **5** favorable numbers.

---

### 🔢 Step 4: Compute the Probability

We calculate the probability as:
$$
\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{1005}
$$

Simplify the fraction:
- GCD of 5 and 1005 is 5
- $ \frac{5}{1005} = \frac{1}{201} $

So, the fraction in lowest terms is:
$$
\frac{a}{b} = \frac{1}{201}
$$

---

### 🔢 Step 5: Compute $ a + b $

From the reduced fraction, we have:
- $ a = 1 $
- $ b = 201 $

Therefore:
$$
a + b = 1 + 201 = \boxed{202}
$$

Solution:
There are 2009 positive integers less than 2010, of which 1005 are odd. If $\frac{1}{n}$ is equal to a terminating decimal, then $n$ can only be divisible by 2 and 5. However, since we have the added restriction that $n$ is odd, $n$ must be a power of 5. There are five powers of 5 less than 2010. \begin{align*}
5^0 &= 1 \\
5^1 &= 5 \\
5^2 &= 25 \\
5^3 &= 125 \\
5^4 &= 625
\end{align*} Note that $5^5 = 3125$. Since there are five odd integers that satisfy our desired condition, the desired probability is $\frac{5}{1005} = \frac{1}{201}$. This is in simplest terms, so our answer is $1+201 = \boxed{202}$.


Explanation:
Okay, let's see... The problem says that a shirt's sale price is $14.40, which is 60% off the original price. I need to find the original price. Hmm, okay. Let me think about how discounts work.

So, if something is 60% off, that means the sale price is the original price minus 60% of the original price. In other words, the sale price is 40% of the original price, right? Because 100% - 60% = 40%. So, the $14.40 is 40% of the original price.

Let me write that down as an equation. Let me call the original price "P". Then, 40% of P equals $14.40.

In mathematical terms, that would be 0.40 * P = 14.40.

So, to find P, I need to divide both sides of the equation by 0.40.

So, P = 14.40 / 0.40.

Let me compute that. Hmm, dividing by 0.40... Well, dividing by 0.4 is the same as multiplying by 2.5, right? Because 1/0.4 is 2.5. Let me check: 0.4 times 2 is 0.8, and 0.4 times 2.5 is 1.0. Yes, that's correct. So, 14.40 divided by 0.40 is the same as 14.40 multiplied by 2.5.

Let me do that multiplication. 14.40 * 2.5. Let me break it down. 14.40 * 2 is 28.80, and 14.40 * 0.5 is 7.20. Adding those together, 28.80 + 7.20 is 36.00.

So, the original price was $36.00? Let me verify that. If the original price is $36, then 60% off would be 0.60 * 36 = $21.60. Subtracting that from $36 gives $36 - $21.60 = $14.40, which matches the sale price given in the problem.

Wait, but just to make sure I didn't make a mistake in my calculation. Let me try another way. If 40% of the original price is $14.40, then 10% of the original price would be $14.40 divided by 4, which is $3.60. Then, 100% would be $3.60 multiplied by 10, which is $36.00. Yep, same answer.

So, that seems consistent. Therefore, the original price must have been $36.

But let me just think again if I interpreted the discount correctly. Sometimes people confuse whether the discount is taken off the original price or something else. But the problem says it's 60% off the original price, so that should mean that the sale price is original price minus 60% of original price, which is 40% of original price. So, that's correct.

Alternatively, if someone thought that 60% off meant that the sale price is 60% of the original price, but that would be incorrect. Because "60% off" means you're subtracting 60% from the original price. So, if it was 60% of the original price, it would be stated as "sale price is 60% of the original price" or "40% off". So, the way it's phrased, "60% off" means subtracting 60%, so the remaining is 40%.

So, that's correct. Therefore, the original price is $36.

I think that's solid. Both methods give the same answer, and checking by reversing the calculation also works. So, I think that's the right answer.

**Final Answer**
The original price of the shirt was \boxed{36} dollars.

To find the original price of the shirt, we analyze the relationship between the sale price and the original price, given that the shirt is sold at **60% off** its original price.

---

### Step 1: Understand the Discount

A **60% off** discount means that the **sale price** is **40% of the original price**, since:

$$
100\% - 60\% = 40\%
$$

So, the **sale price** ($14.40) represents **40%** of the original price.

Let the original price be $ P $.

We can express this relationship mathematically as:

$$
0.40 \times P = 14.40
$$

---

### Step 2: Solve for the Original Price

Now, solve the equation for $ P $ by dividing both sides by 0.40:

$$
P = \frac{14.40}{0.40}
$$

$$
P = 36.00
$$

---

### Step 3: Verify the Result

To confirm, check if 60% off of $36.00 results in the sale price of $14.40:

$$
60\% \text{ of } 36.00 = 0.60 \times 36.00 = 21.60
$$

Now subtract this discount from the original price:

$$
36.00 - 21.60 = 14.40
$$

This matches the given sale price, confirming the accuracy of the calculation.

---

### Final Answer

$$
\boxed{36}
$$

The original price of the shirt was **$36.00**.

[Assistant]
Explanation:
Okay, let's see... The problem says that a shirt's sale price is $14.40, which is 60% off the original price. I need to find the original price. Hmm, okay. Let me think about how discounts work.

So, if something is 60% off, that means the sale price is the original price minus 60% of the original price. In other words, the sale price is 40% of the original price, right? Because 100% - 60% = 40%. So, the $14.40 is 40% of the original price.

Let me write that down as an equation. Let me call the original price "P". Then, 40% of P equals $14.40.

In mathematical terms, that would be 0.40 * P = 14.40.

So, to find P, I need to divide both sides of the equation by 0.40.

So, P = 14.40 / 0.40.

Let me compute that. Hmm, dividing by 0.40... Well, dividing by 0.4 is the same as multiplying by 2.5, right? Because 1/0.4 is 2.5. Let me check: 0.4 times 2 is 0.8, and 0.4 times 2.5 is 1.0. Yes, that's correct. So, 14.40 divided by 0.40 is the same as 14.40 multiplied by 2.5.

Let me do that multiplication. 14.40 * 2.5. Let me break it down. 14.40 * 2 is 28.80, and 14.40 * 0.5 is 7.20. Adding those together, 28.80 + 7.20 is 36.00.

So, the original price was $36.00? Let me verify that. If the original price is $36, then 60% off would be 0.60 * 36 = $21.60. Subtracting that from $36 gives $36 - $21.60 = $14.40, which matches the sale price given in the problem.

Wait, but just to make sure I didn't make a mistake in my calculation. Let me try another way. If 40% of the original price is $14.40, then 10% of the original price would be $14.40 divided by 4, which is $3.60. Then, 100% would be $3.60 multiplied by 10, which is $36.00. Yep, same answer.

So, that seems consistent. Therefore, the original price must have been $36.

But let me just think again if I interpreted the discount correctly. Sometimes people confuse whether the discount is taken off the original price or something else. But the problem says it's 60% off the original price, so that should mean that the sale price is original price minus 60% of original price, which is 40% of original price. So, that's correct.

Alternatively, if someone thought that 60% off meant that the sale price is 60% of the original price, but that would be incorrect. Because "60% off" means you're subtracting 60% from the original price. So, if it was 60% of the original price, it would be stated as "sale price is 60% of the original price" or "40% off". So, the way it's phrased, "60% off" means subtracting 60%, so the remaining is 40%.

So, that's correct. Therefore, the original price is $36.

I think that's solid. Both methods give the same answer, and checking by reversing the calculation also works. So, I think that's the right answer.

**Final Answer**
The original price of the shirt was \boxed{36} dollars.

To find the original price of the shirt, we analyze the relationship between the sale price and the original price, given that the shirt is sold at **60% off** its original price.

---

### Step 1: Understand the Discount

A **60% off** discount means that the **sale price** is **40% of the original price**, since:

$$
100\% - 60\% = 40\%
$$

So, the **sale price** ($14.40) represents **40%** of the original price.

Let the original price be $ P $.

We can express this relationship mathematically as:

$$
0.40 \times P = 14.40
$$

---

### Step 2: Solve for the Original Price

Now, solve the equation for $ P $ by dividing both sides by 0.40:

$$
P = \frac{14.40}{0.40}
$$

$$
P = 36.00
$$

---

### Step 3: Verify the Result

To confirm, check if 60% off of $36.00 results in the sale price of $14.40:

$$
60\% \text{ of } 36.00 = 0.60 \times 36.00 = 21.60
$$

Now subtract this discount from the original price:

$$
36.00 - 21.60 = 14.40
$$

This matches the given sale price, confirming the accuracy of the calculation.

---

### Final Answer

$$
\boxed{36}
$$

The original price of the shirt was **$36.00**.

Solution:
If the shirt is $60\%$ off, it is currently $.4$ of the original price.  Thus the original price was

$$\frac{\$14.40}{.4}=\boxed{\$36}$$