Explanation:
Okay, so I need to figure out what kind of curve is defined by the equation $ r^2 \cos 2\theta = 4 $. The options are Line, Circle, Parabola, Ellipse, or Hyperbola. Let me start by recalling some polar equations and their corresponding curves.

First, I remember that polar equations can represent various conic sections, but I need to recall the specific forms. Also, the equation given here has $ r^2 $ and $ \cos 2\theta $, which makes me think it might be related to some kind of conic section or maybe a different curve. Let me try to manipulate the equation to see if I can convert it into Cartesian coordinates, which might be easier to identify.

So, the original equation is $ r^2 \cos 2\theta = 4 $. Let me recall that in polar coordinates, $ r^2 = x^2 + y^2 $, and $ \cos 2\theta $ can be expressed in terms of x and y. Also, there's a trigonometric identity for $ \cos 2\theta $: $ \cos 2\theta = \cos^2 \theta - \sin^2 \theta $, or alternatively, $ \cos 2\theta = 1 - 2\sin^2 \theta $, or $ \cos 2\theta = 2\cos^2 \theta - 1 $. But maybe there's a better way to express $ \cos 2\theta $ in terms of x and y?

Alternatively, I remember that $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $. Let me check that. Since $ x = r \cos \theta $ and $ y = r \sin \theta $, so $ x^2 = r^2 \cos^2 \theta $, $ y^2 = r^2 \sin^2 \theta $. Therefore, $ x^2 - y^2 = r^2 (\cos^2 \theta - \sin^2 \theta) = r^2 \cos 2\theta $. Therefore, $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $.

So substituting back into the original equation: $ r^2 \times \frac{x^2 - y^2}{r^2} = 4 $. Wait, that simplifies to $ x^2 - y^2 = 4 $.

Oh! So the equation simplifies to $ x^2 - y^2 = 4 $. That's a hyperbola! Because the standard form of a hyperbola is $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $. In this case, if I divide both sides by 4, I get $ \frac{x^2}{4} - \frac{y^2}{4} = 1 $, which is indeed a hyperbola with a = 2 and b = 2.

Wait, but let me make sure I didn't make a mistake in the conversion. Let me go through the steps again. Original equation: $ r^2 \cos 2\theta = 4 $. Then, using the identity $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $, so substituting that into the equation gives $ r^2 \times \frac{x^2 - y^2}{r^2} = 4 $, which simplifies to $ x^2 - y^2 = 4 $. Yes, that seems correct.

Alternatively, maybe I can check by converting the original polar equation to Cartesian coordinates in another way. Let me recall that $ \cos 2\theta $ can also be expressed as $ \frac{r^2 \cos 2\theta}{r^2} $, but that seems redundant. Wait, maybe I can use another identity.

Alternatively, maybe I can use the double angle formula for cosine. Let me think. If I have $ r^2 \cos 2\theta = 4 $, and I know that $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $, so substituting that gives $ r^2 \times \frac{x^2 - y^2}{r^2} = x^2 - y^2 = 4 $. Yep, same result.

So, the Cartesian equation is $ x^2 - y^2 = 4 $, which is a hyperbola. Therefore, the answer should be (E) Hyperbola.

But let me just make sure that there isn't a different interpretation. For example, sometimes polar equations can represent different curves depending on the form. Let me think if there's another way to interpret $ r^2 \cos 2\theta = 4 $.

Alternatively, maybe I can think about the general form of polar equations for conic sections. The standard form for a conic in polar coordinates is $ r = \frac{ed}{1 + e \cos \theta} $ or something similar, where e is the eccentricity. But this equation is different because it's $ r^2 \cos 2\theta = 4 $, which isn't in the standard conic form. However, since we converted it to Cartesian coordinates and found a hyperbola, that seems solid.

Alternatively, maybe I can check specific points. For example, if θ = 0, then cos 0 = 1, so the equation becomes $ r^2 * 1 = 4 $, so r^2 = 4, so r = ±2. So in Cartesian coordinates, when θ = 0, the points are (r, θ) = (2, 0) and (-2, 0), which correspond to (2, 0) and (2, π) which is (-2, 0). So those points are (2, 0) and (-2, 0).

Similarly, if θ = π/4, then cos(2θ) = cos(π/2) = 0, so the equation becomes 0 = 4, which is impossible. Therefore, there are no points at θ = π/4.

If θ = π/2, then cos(2θ) = cos(π) = -1, so the equation becomes r^2 * (-1) = 4 => r^2 = -4, which is impossible. So no points there.

If θ = π/3, then 2θ = 2π/3, cos(2π/3) = -1/2. Therefore, equation becomes r^2 * (-1/2) = 4 => r^2 = -8, which is also impossible.

Wait, so maybe there are only points where cos 2θ is positive? Because otherwise, r^2 would have to be negative, which isn't possible. Therefore, cos 2θ must be positive, so 2θ must be in the first or fourth quadrants, meaning θ is between -π/4 to π/4 or between 3π/4 to 5π/4, etc.

But regardless, the Cartesian equation is x² - y² = 4, which is a hyperbola. So even though in polar coordinates, the equation might have restrictions on θ, the overall shape is a hyperbola.

Alternatively, maybe I can think about the general equation of a hyperbola in polar coordinates. But I think converting to Cartesian coordinates is the straightforward way here.

Another way to check is to recall that the equation x² - y² = 4 is a hyperbola centered at the origin, opening along the x-axis. So that's a standard hyperbola. Therefore, the answer is (E) Hyperbola.

Wait, but let me check if there's a possibility that the original equation could represent something else. For example, sometimes equations can be degenerate or have different forms. But in this case, x² - y² = 4 is a standard hyperbola, not degenerate.

Alternatively, maybe I can think about the polar equation. Let me try to write it in terms of r. Let me rearrange the original equation: $ r^2 \cos 2\theta = 4 $, so $ r^2 = \frac{4}{\cos 2\theta} $. But $ \cos 2\theta $ can be positive or negative. However, since r² is always non-negative, then $ \cos 2\theta $ must be positive. Therefore, the equation is only defined when $ \cos 2\theta > 0 $, which as I thought before, restricts θ to certain regions.

But even so, the Cartesian equation is still x² - y² = 4, which is a hyperbola. Therefore, regardless of the polar form, the curve is a hyperbola.

Therefore, the answer should be (E) Hyperbola.

But just to make sure, let me check if there's any chance that this could be a different conic. For example, if the equation was $ r^2 \cos 2\theta = 4 $, and if I think about the general form of conic sections in polar coordinates. But I think that's more complicated.

Alternatively, maybe I can think of the equation $ r^2 \cos 2\theta = 4 $ as a quadratic equation in polar coordinates. But since converting to Cartesian coordinates gives a hyperbola, which is a conic section, and since hyperbola is one of the options, that's probably the answer.

Alternatively, if I think about the general equation of a hyperbola in polar coordinates. For example, if you have a hyperbola with a horizontal transverse axis, its polar equation can be written as $ r^2 = \frac{b^2}{1 - e^2 \cos^2 \theta} $ or something like that? Wait, maybe not. Let me recall that the standard polar form for conic sections is $ r = \frac{ed}{1 + e \cos \theta} $, but that's for a conic with one focus at the origin. However, the equation given here is different.

Alternatively, maybe I can think of the equation $ x^2 - y^2 = 4 $ as a hyperbola. Since it's a quadratic equation with x² and y² terms having opposite signs, that's the definition of a hyperbola. So that's definitely a hyperbola.

Therefore, I think I can confidently say that the answer is (E) Hyperbola.

**Final Answer**
\boxed{E}

To determine the type of curve defined by the polar equation

$$
r^2 \cos 2\theta = 4,
$$

we begin by converting it into Cartesian coordinates, which is often more intuitive for identifying conic sections.

---

### Step 1: Use Trigonometric Identity

Recall the identity:

$$
\cos 2\theta = \frac{x^2 - y^2}{r^2}.
$$

This identity comes from the fact that:

$$
x = r \cos \theta, \quad y = r \sin \theta \Rightarrow x^2 + y^2 = r^2.
$$

So,

$$
x^2 - y^2 = r^2 (\cos^2 \theta - \sin^2 \theta) = r^2 \cos 2\theta.
$$

Therefore, we can substitute this into the original equation:

$$
r^2 \cos 2\theta = 4 \Rightarrow r^2 \cdot \frac{x^2 - y^2}{r^2} = 4.
$$

Simplifying:

$$
x^2 - y^2 = 4.
$$

---

### Step 2: Identify the Curve

The equation $ x^2 - y^2 = 4 $ is a standard form of a **hyperbola**. To see this more clearly, divide both sides by 4:

$$
\frac{x^2}{4} - \frac{y^2}{4} = 1.
$$

This is the standard form of a hyperbola centered at the origin, opening along the x-axis, with:

- $ a^2 = 4 \Rightarrow a = 2 $,
- $ b^2 = 4 \Rightarrow b = 2 $.

---

### Step 3: Confirm the Nature of the Curve

The original polar equation is defined only when $ \cos 2\theta > 0 $, since $ r^2 $ must be non-negative. This restricts the values of $ \theta $ to regions where $ \cos 2\theta > 0 $, which corresponds to the branches of the hyperbola in the Cartesian plane.

Thus, the curve is a **hyperbola**, and the correct answer is:

$$
\boxed{E}
$$

[Assistant]
Explanation:
Okay, so I need to figure out what kind of curve is defined by the equation $ r^2 \cos 2\theta = 4 $. The options are Line, Circle, Parabola, Ellipse, or Hyperbola. Let me start by recalling some polar equations and their corresponding curves.

First, I remember that polar equations can represent various conic sections, but I need to recall the specific forms. Also, the equation given here has $ r^2 $ and $ \cos 2\theta $, which makes me think it might be related to some kind of conic section or maybe a different curve. Let me try to manipulate the equation to see if I can convert it into Cartesian coordinates, which might be easier to identify.

So, the original equation is $ r^2 \cos 2\theta = 4 $. Let me recall that in polar coordinates, $ r^2 = x^2 + y^2 $, and $ \cos 2\theta $ can be expressed in terms of x and y. Also, there's a trigonometric identity for $ \cos 2\theta $: $ \cos 2\theta = \cos^2 \theta - \sin^2 \theta $, or alternatively, $ \cos 2\theta = 1 - 2\sin^2 \theta $, or $ \cos 2\theta = 2\cos^2 \theta - 1 $. But maybe there's a better way to express $ \cos 2\theta $ in terms of x and y?

Alternatively, I remember that $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $. Let me check that. Since $ x = r \cos \theta $ and $ y = r \sin \theta $, so $ x^2 = r^2 \cos^2 \theta $, $ y^2 = r^2 \sin^2 \theta $. Therefore, $ x^2 - y^2 = r^2 (\cos^2 \theta - \sin^2 \theta) = r^2 \cos 2\theta $. Therefore, $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $.

So substituting back into the original equation: $ r^2 \times \frac{x^2 - y^2}{r^2} = 4 $. Wait, that simplifies to $ x^2 - y^2 = 4 $.

Oh! So the equation simplifies to $ x^2 - y^2 = 4 $. That's a hyperbola! Because the standard form of a hyperbola is $ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 $. In this case, if I divide both sides by 4, I get $ \frac{x^2}{4} - \frac{y^2}{4} = 1 $, which is indeed a hyperbola with a = 2 and b = 2.

Wait, but let me make sure I didn't make a mistake in the conversion. Let me go through the steps again. Original equation: $ r^2 \cos 2\theta = 4 $. Then, using the identity $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $, so substituting that into the equation gives $ r^2 \times \frac{x^2 - y^2}{r^2} = 4 $, which simplifies to $ x^2 - y^2 = 4 $. Yes, that seems correct.

Alternatively, maybe I can check by converting the original polar equation to Cartesian coordinates in another way. Let me recall that $ \cos 2\theta $ can also be expressed as $ \frac{r^2 \cos 2\theta}{r^2} $, but that seems redundant. Wait, maybe I can use another identity.

Alternatively, maybe I can use the double angle formula for cosine. Let me think. If I have $ r^2 \cos 2\theta = 4 $, and I know that $ \cos 2\theta = \frac{x^2 - y^2}{r^2} $, so substituting that gives $ r^2 \times \frac{x^2 - y^2}{r^2} = x^2 - y^2 = 4 $. Yep, same result.

So, the Cartesian equation is $ x^2 - y^2 = 4 $, which is a hyperbola. Therefore, the answer should be (E) Hyperbola.

But let me just make sure that there isn't a different interpretation. For example, sometimes polar equations can represent different curves depending on the form. Let me think if there's another way to interpret $ r^2 \cos 2\theta = 4 $.

Alternatively, maybe I can think about the general form of polar equations for conic sections. The standard form for a conic in polar coordinates is $ r = \frac{ed}{1 + e \cos \theta} $ or something similar, where e is the eccentricity. But this equation is different because it's $ r^2 \cos 2\theta = 4 $, which isn't in the standard conic form. However, since we converted it to Cartesian coordinates and found a hyperbola, that seems solid.

Alternatively, maybe I can check specific points. For example, if θ = 0, then cos 0 = 1, so the equation becomes $ r^2 * 1 = 4 $, so r^2 = 4, so r = ±2. So in Cartesian coordinates, when θ = 0, the points are (r, θ) = (2, 0) and (-2, 0), which correspond to (2, 0) and (2, π) which is (-2, 0). So those points are (2, 0) and (-2, 0).

Similarly, if θ = π/4, then cos(2θ) = cos(π/2) = 0, so the equation becomes 0 = 4, which is impossible. Therefore, there are no points at θ = π/4.

If θ = π/2, then cos(2θ) = cos(π) = -1, so the equation becomes r^2 * (-1) = 4 => r^2 = -4, which is impossible. So no points there.

If θ = π/3, then 2θ = 2π/3, cos(2π/3) = -1/2. Therefore, equation becomes r^2 * (-1/2) = 4 => r^2 = -8, which is also impossible.

Wait, so maybe there are only points where cos 2θ is positive? Because otherwise, r^2 would have to be negative, which isn't possible. Therefore, cos 2θ must be positive, so 2θ must be in the first or fourth quadrants, meaning θ is between -π/4 to π/4 or between 3π/4 to 5π/4, etc.

But regardless, the Cartesian equation is x² - y² = 4, which is a hyperbola. So even though in polar coordinates, the equation might have restrictions on θ, the overall shape is a hyperbola.

Alternatively, maybe I can think about the general equation of a hyperbola in polar coordinates. But I think converting to Cartesian coordinates is the straightforward way here.

Another way to check is to recall that the equation x² - y² = 4 is a hyperbola centered at the origin, opening along the x-axis. So that's a standard hyperbola. Therefore, the answer is (E) Hyperbola.

Wait, but let me check if there's a possibility that the original equation could represent something else. For example, sometimes equations can be degenerate or have different forms. But in this case, x² - y² = 4 is a standard hyperbola, not degenerate.

Alternatively, maybe I can think about the polar equation. Let me try to write it in terms of r. Let me rearrange the original equation: $ r^2 \cos 2\theta = 4 $, so $ r^2 = \frac{4}{\cos 2\theta} $. But $ \cos 2\theta $ can be positive or negative. However, since r² is always non-negative, then $ \cos 2\theta $ must be positive. Therefore, the equation is only defined when $ \cos 2\theta > 0 $, which as I thought before, restricts θ to certain regions.

But even so, the Cartesian equation is still x² - y² = 4, which is a hyperbola. Therefore, regardless of the polar form, the curve is a hyperbola.

Therefore, the answer should be (E) Hyperbola.

But just to make sure, let me check if there's any chance that this could be a different conic. For example, if the equation was $ r^2 \cos 2\theta = 4 $, and if I think about the general form of conic sections in polar coordinates. But I think that's more complicated.

Alternatively, maybe I can think of the equation $ r^2 \cos 2\theta = 4 $ as a quadratic equation in polar coordinates. But since converting to Cartesian coordinates gives a hyperbola, which is a conic section, and since hyperbola is one of the options, that's probably the answer.

Alternatively, if I think about the general equation of a hyperbola in polar coordinates. For example, if you have a hyperbola with a horizontal transverse axis, its polar equation can be written as $ r^2 = \frac{b^2}{1 - e^2 \cos^2 \theta} $ or something like that? Wait, maybe not. Let me recall that the standard polar form for conic sections is $ r = \frac{ed}{1 + e \cos \theta} $, but that's for a conic with one focus at the origin. However, the equation given here is different.

Alternatively, maybe I can think of the equation $ x^2 - y^2 = 4 $ as a hyperbola. Since it's a quadratic equation with x² and y² terms having opposite signs, that's the definition of a hyperbola. So that's definitely a hyperbola.

Therefore, I think I can confidently say that the answer is (E) Hyperbola.

**Final Answer**
\boxed{E}

To determine the type of curve defined by the polar equation

$$
r^2 \cos 2\theta = 4,
$$

we begin by converting it into Cartesian coordinates, which is often more intuitive for identifying conic sections.

---

### Step 1: Use Trigonometric Identity

Recall the identity:

$$
\cos 2\theta = \frac{x^2 - y^2}{r^2}.
$$

This identity comes from the fact that:

$$
x = r \cos \theta, \quad y = r \sin \theta \Rightarrow x^2 + y^2 = r^2.
$$

So,

$$
x^2 - y^2 = r^2 (\cos^2 \theta - \sin^2 \theta) = r^2 \cos 2\theta.
$$

Therefore, we can substitute this into the original equation:

$$
r^2 \cos 2\theta = 4 \Rightarrow r^2 \cdot \frac{x^2 - y^2}{r^2} = 4.
$$

Simplifying:

$$
x^2 - y^2 = 4.
$$

---

### Step 2: Identify the Curve

The equation $ x^2 - y^2 = 4 $ is a standard form of a **hyperbola**. To see this more clearly, divide both sides by 4:

$$
\frac{x^2}{4} - \frac{y^2}{4} = 1.
$$

This is the standard form of a hyperbola centered at the origin, opening along the x-axis, with:

- $ a^2 = 4 \Rightarrow a = 2 $,
- $ b^2 = 4 \Rightarrow b = 2 $.

---

### Step 3: Confirm the Nature of the Curve

The original polar equation is defined only when $ \cos 2\theta > 0 $, since $ r^2 $ must be non-negative. This restricts the values of $ \theta $ to regions where $ \cos 2\theta > 0 $, which corresponds to the branches of the hyperbola in the Cartesian plane.

Thus, the curve is a **hyperbola**, and the correct answer is:

$$
\boxed{E}
$$

Solution:
From $r^2 \cos 2 \theta = 4,$
\[r^2 (\cos^2 \theta - \sin^2 \theta) = r^2 \cos^2 \theta - r^2 \sin^2 \theta = 4.\]Then $x^2 - y^2 = 4,$ or
\[\frac{x^2}{4} - \frac{y^2}{4} = 1.\]Thus, the graph represents a hyperbola.  The answer is $\boxed{\text{(E)}}.$

[asy]
unitsize(0.5 cm);

pair moo (real t) {
  real r = sqrt(4/Cos(2*t));
  return (r*Cos(t), r*Sin(t));
}

path foo = moo(-44);
real t;

for (t = -44; t <= 44; t = t + 0.1) {
  foo = foo--moo(t);
}

draw(foo,red);
draw(reflect((0,0),(0,1))*(foo),red);

draw((-4,0)--(4,0));
draw((0,-4)--(0,4));

limits((-4,-4),(4,4),Crop);

label("$r^2 \cos 2 \theta = 4$", (6.5,1.5), red);
[/asy]


Explanation:
Okay, so I need to find the 2003rd term of the sequence of odd numbers: 1, 3, 5, 7, and so on. Let me think about how sequences like this work.

First, I remember that the sequence of odd numbers is an arithmetic sequence. That means each term increases by a constant difference. In this case, the difference between each term is 2. Let me confirm that: 3 - 1 = 2, 5 - 3 = 2, 7 - 5 = 2... yeah, that's right. So the common difference (d) is 2.

Now, arithmetic sequences have a formula for the nth term. I think the formula is something like a_n = a_1 + (n - 1)d, where a_n is the nth term, a_1 is the first term, d is the common difference, and n is the term number. Let me check if I remember that correctly.

Let me test it with the first few terms. For n = 1, a_1 = 1 + (1 - 1)*2 = 1 + 0 = 1. That's correct. For n = 2, a_2 = 1 + (2 - 1)*2 = 1 + 2 = 3. Yep, that works. For n = 3, 1 + (3 - 1)*2 = 1 + 4 = 5. Perfect. So the formula seems right.

So applying this formula to find the 2003rd term. Let me write that down:

a_2003 = a_1 + (2003 - 1)*d

We know a_1 is 1, d is 2. So plugging those in:

a_2003 = 1 + (2002)*2

Let me compute that. 2002 multiplied by 2. Well, 2000*2 is 4000, and 2*2 is 4, so total is 4000 + 4 = 4004. Then add the 1 from the original formula:

a_2003 = 1 + 4004 = 4005.

Wait, so is the answer 4005? Let me check again to make sure I didn't make a mistake.

Alternatively, maybe there's another way to think about odd numbers. I remember that the nth odd number is given by the formula 2n - 1. Let me verify that. For n = 1, 2*1 - 1 = 1. For n = 2, 2*2 - 1 = 3. For n = 3, 2*3 - 1 = 5. Yep, that works. So if that's the case, then the 2003rd term would be 2*2003 - 1. Let me calculate that.

2*2003 is 4006, minus 1 is 4005. So same answer. Okay, so that seems to confirm it.

But wait, why is there a difference between the two formulas? Wait, no, actually, they are the same. Let me check. The formula from the arithmetic sequence is a_n = a_1 + (n - 1)d. Here, a_1 is 1, d is 2, so:

a_n = 1 + (n - 1)*2 = 1 + 2n - 2 = 2n - 1.

Oh, right! So they are the same formula. So both methods lead to the same result. Therefore, the 2003rd term is 2*2003 - 1 = 4005.

But just to be thorough, let me think of another way. Maybe by considering how odd numbers are structured.

Odd numbers can be thought of as numbers that are not divisible by 2. So, the first odd number is 1, then 3, 5, etc. Each odd number is 1 more than an even number. But maybe that's complicating things. Alternatively, if I think about the position of the term.

For example, the first term is 1, which is 2*1 - 1. The second term is 3, which is 2*2 - 1. So, in general, the nth term is 2n - 1. Therefore, the 2003rd term is 2*2003 - 1.

But let me think if there's a possibility that the sequence starts at n=0? Wait, no. The sequence is given as 1, 3, 5, 7,... so the first term is 1, which is n=1. So the formula 2n - 1 is correct here.

Alternatively, if someone thought that the sequence starts at n=0, then the 0th term would be 1, but that's not the case here. The problem says "the sequence of odd numbers 1, 3, 5, 7, …", so the first term is 1, so n starts at 1.

Another way to think about it: if you have the sequence of natural numbers, which is 1, 2, 3, 4, 5,... and you take the odd ones, so every other number. So the nth odd number is the number at position 2n - 1 in the natural numbers. Wait, let me check that.

For example, the first odd number is 1, which is the 1st natural number. The second odd number is 3, which is the 2nd odd number, but in natural numbers, it's the 3rd number. Wait, so maybe the nth odd number is the (2n - 1)th natural number? Wait, that would be 2n - 1. So that's the same formula again.

So, for example, the 1st odd number is 2*1 - 1 = 1, the 2nd is 2*2 - 1 = 3, the 3rd is 5, etc. So that's consistent. Therefore, the 2003rd term is 2*2003 - 1 = 4005.

Alternatively, if I think about the sum of the first n odd numbers, which is n². Wait, that's another formula. The sum of the first n odd numbers is n squared. But that's a different concept. However, maybe that can help me check?

Wait, if I know that the sum of the first n odd numbers is n², but how does that relate to the nth term? Well, the nth term is the difference between the sum of the first n terms and the sum of the first (n-1) terms.

So, sum of first n odd numbers is n², sum of first (n-1) is (n - 1)². Therefore, the nth term is n² - (n - 1)². Let me compute that:

n² - (n - 1)² = n² - (n² - 2n + 1) = n² - n² + 2n - 1 = 2n - 1.

So again, same formula. Therefore, the nth term is 2n - 1. Therefore, the 2003rd term is 2*2003 - 1 = 4005.

So all these different approaches lead me to the same answer. Therefore, I can be confident that the 2003rd term is 4005.

But just to make sure, let me try with smaller numbers. Let's say n=1: 2*1 -1=1, correct. n=2: 3, correct. n=3: 5, correct. n=4: 7, correct. So that works.

Alternatively, if I use the arithmetic sequence formula. For n=1: 1 + (1-1)*2=1. For n=2: 1 + (2-1)*2=3. For n=3: 1 + 2*2=5. Yep. So that works.

Therefore, applying the formula for n=2003: 1 + (2003 -1)*2 = 1 + 2002*2.

Calculating 2002*2: 2000*2=4000, 2*2=4, so total 4004. Then add 1: 4004 +1=4005.

So that's the same answer.

Alternatively, if I think about how many numbers are there between 1 and 4005. Wait, but maybe that's not necessary.

Alternatively, think of the odd numbers as 1, 3, 5,... which can be represented as 2k -1 where k is 1,2,3,... So if I want the 2003rd term, then k=2003, so term is 2*2003 -1=4005.

So all methods converge to 4005.

Therefore, I think the answer is 4005.

But just to make sure, let me check with another approach. Let's think about the position of the term.

Each odd number can be associated with its position. For example, term 1 is 1, term 2 is 3, term 3 is 5, term 4 is 7, etc. So each term is 2 more than the previous term. So starting at 1, adding 2 each time.

So the sequence is 1, 1+2, 1+2+2, 1+2+2+2,... Wait, no. Wait, the nth term is 1 + 2*(n-1). Which is the same as 2n -1.

Yes, because 1 + 2*(n -1) = 1 + 2n - 2 = 2n -1.

So again, same formula.

Therefore, regardless of how I approach it, the formula is 2n -1.

Therefore, 2003rd term is 2*2003 -1.

Calculating 2*2003: 2000*2=4000, 3*2=6, so total 4006. Then subtract 1: 4006 -1=4005.

Yes.

Alternatively, if I have a calculator, I can compute 2003*2. But since I don't have a calculator here, I can do it manually.

2003 * 2:

Breakdown: 2000*2 + 3*2 = 4000 + 6 = 4006.

Subtract 1: 4006 -1 = 4005.

So that's correct.

Therefore, I think I can be confident that the answer is 4005.

But just to make sure, let me think if there's any chance that the question is referring to something else. For example, sometimes sequences can be defined differently, but the problem says "the sequence of odd numbers 1, 3, 5, 7, …". So that's the standard sequence of odd numbers, starting at 1, increasing by 2 each time.

Therefore, the formula applies.

Alternatively, if someone thought that the sequence starts at 0, but no, the first term is 1.

Alternatively, if someone thought that the sequence is 1, 3, 5,... but starting from n=0, but again, the problem says the first term is 1, so n=1.

Therefore, no issues there.

Another way: think of the odd numbers as all numbers congruent to 1 mod 2. So the nth term is 1 + 2*(n-1). Which again is 2n -1.

Therefore, same result.

Alternatively, think of the odd numbers as the numbers that are not divisible by 2. So, the nth odd number is the number that is 2n -1.

Therefore, again, same answer.

So, after all these different ways of thinking about it, I can't find any mistake in my reasoning. Therefore, I think the answer is definitely 4005.

**Final Answer**
The 2003rd term of the sequence is \boxed{4005}.

To determine the 2003rd term of the sequence of odd numbers:
$$
1, 3, 5, 7, \dots
$$
we recognize that this is an **arithmetic sequence** with the following properties:

- **First term** $ a_1 = 1 $
- **Common difference** $ d = 2 $

---

### Step 1: Use the General Formula for the $ n $-th Term of an Arithmetic Sequence

The general formula for the $ n $-th term of an arithmetic sequence is:

$$
a_n = a_1 + (n - 1)d
$$

Substituting the known values:

$$
a_{2003} = 1 + (2003 - 1) \cdot 2 = 1 + 2002 \cdot 2
$$

$$
a_{2003} = 1 + 4004 = 4005
$$

---

### Step 2: Confirm with an Alternative Formula

There is a well-known formula for the $ n $-th odd number:

$$
a_n = 2n - 1
$$

Applying this directly:

$$
a_{2003} = 2 \cdot 2003 - 1 = 4006 - 1 = 4005
$$

This confirms the result.

---

### Final Answer

$$
\boxed{4005}
$$

[Assistant]
Explanation:
Okay, so I need to find the 2003rd term of the sequence of odd numbers: 1, 3, 5, 7, and so on. Let me think about how sequences like this work.

First, I remember that the sequence of odd numbers is an arithmetic sequence. That means each term increases by a constant difference. In this case, the difference between each term is 2. Let me confirm that: 3 - 1 = 2, 5 - 3 = 2, 7 - 5 = 2... yeah, that's right. So the common difference (d) is 2.

Now, arithmetic sequences have a formula for the nth term. I think the formula is something like a_n = a_1 + (n - 1)d, where a_n is the nth term, a_1 is the first term, d is the common difference, and n is the term number. Let me check if I remember that correctly.

Let me test it with the first few terms. For n = 1, a_1 = 1 + (1 - 1)*2 = 1 + 0 = 1. That's correct. For n = 2, a_2 = 1 + (2 - 1)*2 = 1 + 2 = 3. Yep, that works. For n = 3, 1 + (3 - 1)*2 = 1 + 4 = 5. Perfect. So the formula seems right.

So applying this formula to find the 2003rd term. Let me write that down:

a_2003 = a_1 + (2003 - 1)*d

We know a_1 is 1, d is 2. So plugging those in:

a_2003 = 1 + (2002)*2

Let me compute that. 2002 multiplied by 2. Well, 2000*2 is 4000, and 2*2 is 4, so total is 4000 + 4 = 4004. Then add the 1 from the original formula:

a_2003 = 1 + 4004 = 4005.

Wait, so is the answer 4005? Let me check again to make sure I didn't make a mistake.

Alternatively, maybe there's another way to think about odd numbers. I remember that the nth odd number is given by the formula 2n - 1. Let me verify that. For n = 1, 2*1 - 1 = 1. For n = 2, 2*2 - 1 = 3. For n = 3, 2*3 - 1 = 5. Yep, that works. So if that's the case, then the 2003rd term would be 2*2003 - 1. Let me calculate that.

2*2003 is 4006, minus 1 is 4005. So same answer. Okay, so that seems to confirm it.

But wait, why is there a difference between the two formulas? Wait, no, actually, they are the same. Let me check. The formula from the arithmetic sequence is a_n = a_1 + (n - 1)d. Here, a_1 is 1, d is 2, so:

a_n = 1 + (n - 1)*2 = 1 + 2n - 2 = 2n - 1.

Oh, right! So they are the same formula. So both methods lead to the same result. Therefore, the 2003rd term is 2*2003 - 1 = 4005.

But just to be thorough, let me think of another way. Maybe by considering how odd numbers are structured.

Odd numbers can be thought of as numbers that are not divisible by 2. So, the first odd number is 1, then 3, 5, etc. Each odd number is 1 more than an even number. But maybe that's complicating things. Alternatively, if I think about the position of the term.

For example, the first term is 1, which is 2*1 - 1. The second term is 3, which is 2*2 - 1. So, in general, the nth term is 2n - 1. Therefore, the 2003rd term is 2*2003 - 1.

But let me think if there's a possibility that the sequence starts at n=0? Wait, no. The sequence is given as 1, 3, 5, 7,... so the first term is 1, which is n=1. So the formula 2n - 1 is correct here.

Alternatively, if someone thought that the sequence starts at n=0, then the 0th term would be 1, but that's not the case here. The problem says "the sequence of odd numbers 1, 3, 5, 7, …", so the first term is 1, so n starts at 1.

Another way to think about it: if you have the sequence of natural numbers, which is 1, 2, 3, 4, 5,... and you take the odd ones, so every other number. So the nth odd number is the number at position 2n - 1 in the natural numbers. Wait, let me check that.

For example, the first odd number is 1, which is the 1st natural number. The second odd number is 3, which is the 2nd odd number, but in natural numbers, it's the 3rd number. Wait, so maybe the nth odd number is the (2n - 1)th natural number? Wait, that would be 2n - 1. So that's the same formula again.

So, for example, the 1st odd number is 2*1 - 1 = 1, the 2nd is 2*2 - 1 = 3, the 3rd is 5, etc. So that's consistent. Therefore, the 2003rd term is 2*2003 - 1 = 4005.

Alternatively, if I think about the sum of the first n odd numbers, which is n². Wait, that's another formula. The sum of the first n odd numbers is n squared. But that's a different concept. However, maybe that can help me check?

Wait, if I know that the sum of the first n odd numbers is n², but how does that relate to the nth term? Well, the nth term is the difference between the sum of the first n terms and the sum of the first (n-1) terms.

So, sum of first n odd numbers is n², sum of first (n-1) is (n - 1)². Therefore, the nth term is n² - (n - 1)². Let me compute that:

n² - (n - 1)² = n² - (n² - 2n + 1) = n² - n² + 2n - 1 = 2n - 1.

So again, same formula. Therefore, the nth term is 2n - 1. Therefore, the 2003rd term is 2*2003 - 1 = 4005.

So all these different approaches lead me to the same answer. Therefore, I can be confident that the 2003rd term is 4005.

But just to make sure, let me try with smaller numbers. Let's say n=1: 2*1 -1=1, correct. n=2: 3, correct. n=3: 5, correct. n=4: 7, correct. So that works.

Alternatively, if I use the arithmetic sequence formula. For n=1: 1 + (1-1)*2=1. For n=2: 1 + (2-1)*2=3. For n=3: 1 + 2*2=5. Yep. So that works.

Therefore, applying the formula for n=2003: 1 + (2003 -1)*2 = 1 + 2002*2.

Calculating 2002*2: 2000*2=4000, 2*2=4, so total 4004. Then add 1: 4004 +1=4005.

So that's the same answer.

Alternatively, if I think about how many numbers are there between 1 and 4005. Wait, but maybe that's not necessary.

Alternatively, think of the odd numbers as 1, 3, 5,... which can be represented as 2k -1 where k is 1,2,3,... So if I want the 2003rd term, then k=2003, so term is 2*2003 -1=4005.

So all methods converge to 4005.

Therefore, I think the answer is 4005.

But just to make sure, let me check with another approach. Let's think about the position of the term.

Each odd number can be associated with its position. For example, term 1 is 1, term 2 is 3, term 3 is 5, term 4 is 7, etc. So each term is 2 more than the previous term. So starting at 1, adding 2 each time.

So the sequence is 1, 1+2, 1+2+2, 1+2+2+2,... Wait, no. Wait, the nth term is 1 + 2*(n-1). Which is the same as 2n -1.

Yes, because 1 + 2*(n -1) = 1 + 2n - 2 = 2n -1.

So again, same formula.

Therefore, regardless of how I approach it, the formula is 2n -1.

Therefore, 2003rd term is 2*2003 -1.

Calculating 2*2003: 2000*2=4000, 3*2=6, so total 4006. Then subtract 1: 4006 -1=4005.

Yes.

Alternatively, if I have a calculator, I can compute 2003*2. But since I don't have a calculator here, I can do it manually.

2003 * 2:

Breakdown: 2000*2 + 3*2 = 4000 + 6 = 4006.

Subtract 1: 4006 -1 = 4005.

So that's correct.

Therefore, I think I can be confident that the answer is 4005.

But just to make sure, let me think if there's any chance that the question is referring to something else. For example, sometimes sequences can be defined differently, but the problem says "the sequence of odd numbers 1, 3, 5, 7, …". So that's the standard sequence of odd numbers, starting at 1, increasing by 2 each time.

Therefore, the formula applies.

Alternatively, if someone thought that the sequence starts at 0, but no, the first term is 1.

Alternatively, if someone thought that the sequence is 1, 3, 5,... but starting from n=0, but again, the problem says the first term is 1, so n=1.

Therefore, no issues there.

Another way: think of the odd numbers as all numbers congruent to 1 mod 2. So the nth term is 1 + 2*(n-1). Which again is 2n -1.

Therefore, same result.

Alternatively, think of the odd numbers as the numbers that are not divisible by 2. So, the nth odd number is the number that is 2n -1.

Therefore, again, same answer.

So, after all these different ways of thinking about it, I can't find any mistake in my reasoning. Therefore, I think the answer is definitely 4005.

**Final Answer**
The 2003rd term of the sequence is \boxed{4005}.

To determine the 2003rd term of the sequence of odd numbers:
$$
1, 3, 5, 7, \dots
$$
we recognize that this is an **arithmetic sequence** with the following properties:

- **First term** $ a_1 = 1 $
- **Common difference** $ d = 2 $

---

### Step 1: Use the General Formula for the $ n $-th Term of an Arithmetic Sequence

The general formula for the $ n $-th term of an arithmetic sequence is:

$$
a_n = a_1 + (n - 1)d
$$

Substituting the known values:

$$
a_{2003} = 1 + (2003 - 1) \cdot 2 = 1 + 2002 \cdot 2
$$

$$
a_{2003} = 1 + 4004 = 4005
$$

---

### Step 2: Confirm with an Alternative Formula

There is a well-known formula for the $ n $-th odd number:

$$
a_n = 2n - 1
$$

Applying this directly:

$$
a_{2003} = 2 \cdot 2003 - 1 = 4006 - 1 = 4005
$$

This confirms the result.

---

### Final Answer

$$
\boxed{4005}
$$

Solution: The sequence of odd numbers 1, 3, 5, 7, and so on, is an arithmetic sequence, with common difference 2.  Therefore, the $2003^{\text{rd}}$ term is $1+2002\cdot2=\boxed{4005}$.


Explanation:
Okay, let me try to figure out this problem. So, we have two functions, f(x) = 2x - 3 and g(x) = x + 1. The question is asking for the value of g(f(5) - 1). Hmm, let me break this down step by step.

First, I need to find f(5). Since f(x) is defined as 2x - 3, substituting x with 5 should give me the value of f(5). Let me write that out:

f(5) = 2*(5) - 3.

Calculating that, 2 times 5 is 10, and then subtracting 3 gives me 10 - 3, which equals 7. So, f(5) is 7.

Now, the next part of the problem is g(f(5) - 1). Since I already found that f(5) is 7, I can substitute that into the expression. So, inside the parentheses, it's 7 - 1. Let me compute that: 7 minus 1 is 6.

So now, the problem simplifies to finding g(6). Because g(f(5) - 1) becomes g(6).

Now, let me recall what g(x) is. It's given as x + 1. Therefore, substituting x with 6, we have:

g(6) = 6 + 1.

Adding those together, 6 + 1 equals 7.

Wait, so putting it all together, g(f(5) - 1) equals 7? Let me check my steps again to make sure I didn't make a mistake.

First, f(5): 2*5 is 10, minus 3 is 7. That seems right. Then subtract 1 from that result: 7 - 1 is 6. Then plug into g(x): 6 + 1 is 7. Yeah, that seems correct.

But let me think again. Maybe I misread the problem? Let me check the original question again. It says g(f(5) - 1). So, is it g of (f(5) minus 1), or is there a different order of operations? For example, could it be interpreted as g(f(5 - 1))? But no, the parentheses are around f(5) - 1, so it's definitely g evaluated at (f(5) - 1).

Alternatively, maybe I need to consider if there's a different way to parse the expression? Let me confirm. The expression is g(f(5) - 1). Since function composition is usually written with parentheses, but here it's just a function g applied to the result of f(5) minus 1. So, yes, first compute f(5), subtract 1, then apply g.

So, my steps are correct. Let me verify once more:

f(5) = 2*5 - 3 = 10 - 3 = 7.

Then, f(5) - 1 = 7 - 1 = 6.

Then, g(6) = 6 + 1 = 7.

Therefore, the value of g(f(5) - 1) is 7.

Wait, but just to be thorough, let me check if there's another way someone might approach this problem and maybe make a mistake. For example, someone might think that g(f(5) - 1) is the same as g(f(5 - 1)), which would be different. Let me see what that would be. If someone incorrectly interprets the parentheses, they might compute f(5 - 1) first. Let's see:

If it were g(f(5 - 1)), then 5 - 1 is 4, then f(4) is 2*4 - 3 = 8 - 3 = 5, then g(5) is 5 + 1 = 6. But that's different from 7. However, the original problem is g(f(5) - 1), not g(f(5 - 1)). So, the parentheses are important here.

Therefore, as long as we follow the order of operations and the parentheses, we should be okay.

Another possible mistake could be miscalculating f(5). Let me check that again. f(x) = 2x - 3. So, plugging in 5: 2*5 is 10, minus 3 is 7. Correct.

Then subtracting 1: 7 - 1 is 6. Then g(6) is 6 + 1, which is 7.

Alternatively, maybe someone might think that g(f(5) - 1) is g(f(5)) - 1? Let me check that. If that were the case, then g(f(5)) would be g(7) which is 7 + 1 = 8, then subtract 1 would be 7. Wait, so in this case, both interpretations give the same result?

Wait, that's interesting. Let me think. If you have g(f(5) - 1), is that the same as g(f(5)) - 1?

No, not necessarily. Because g is a function, so g(a - b) is not the same as g(a) - b unless the function has specific properties.

But in this case, let me check. If we compute g(f(5) - 1) as per the original problem, we get 7. If we compute g(f(5)) - 1, that would be (g(7)) - 1 = 8 - 1 = 7. So, in this case, both ways give the same answer.

But that's a coincidence? Let me check with different numbers. Suppose f(5) was 8 instead of 7. Then, g(f(5) - 1) would be g(7) = 8. Whereas g(f(5)) - 1 would be g(8) - 1 = 9 - 1 = 8. So, same result. Wait, so is this always true?

Wait, let me think. If g(x) = x + 1, then g(a - 1) = (a - 1) + 1 = a. And g(a) - 1 = (a + 1) - 1 = a. So, in this case, they are equal.

Therefore, in this specific problem, because g(x) is x + 1, then g(f(5) - 1) equals g(f(5)) - 1. So, even if someone misinterprets the parentheses, they might still get the same answer. But that's because of the specific form of g(x).

But in general, this is not true. For example, if g(x) was something else, like g(x) = 2x, then g(a - 1) would be 2(a - 1) = 2a - 2, whereas g(a) - 1 would be 2a - 1. Which are different.

So, in this problem, because of the specific function g(x) = x + 1, both interpretations give the same result. But the correct way is to follow the parentheses.

But since both ways give the same answer here, maybe the problem is designed that way? But regardless, the correct approach is to compute f(5) first, subtract 1, then apply g.

So, in conclusion, the answer is 7.

But just to make sure, let me write all steps again:

1. Compute f(5):

f(5) = 2*(5) - 3 = 10 - 3 = 7.

2. Compute f(5) - 1:

7 - 1 = 6.

3. Compute g(6):

g(6) = 6 + 1 = 7.

Therefore, the value of g(f(5) - 1) is 7.

Yes, that seems consistent.

Alternatively, if I wanted to write this as a composition of functions, maybe? Let me think. But the problem isn't asking for a composition, it's asking for g evaluated at (f(5) - 1). So, no, it's not a composition of functions, it's a function evaluated at an expression.

Alternatively, if it were g(f(x)) - 1, then we would have a different story, but no, the problem is specifically g(f(5) - 1).

Another way to think about it: maybe substitute the functions into each other? Let me try that.

First, let's find f(5) - 1. As we found, that's 7 - 1 = 6. Then, g(6) is 6 + 1 = 7.

Alternatively, if I wanted to write an expression for g(f(x) - 1), I could do that. Let me try:

g(f(x) - 1) = [f(x) - 1] + 1 = f(x) - 1 + 1 = f(x).

Wait, that's interesting. So, if you have g(f(x) - 1), since g is x + 1, then substituting f(x) - 1 into g gives [f(x) - 1] + 1 = f(x). Therefore, in general, g(f(x) - 1) = f(x).

Therefore, if we plug in x = 5, then g(f(5) - 1) = f(5) = 7. Which matches our previous result.

So, that's another way to see it. Since g is the function that adds 1, subtracting 1 before applying g cancels out the addition. Therefore, g(f(x) - 1) simplifies to f(x).

Therefore, regardless of the value of x, g(f(x) - 1) is equal to f(x). Therefore, for x = 5, it's f(5) = 7.

This seems like a more general approach. So, if I can recognize that g is the inverse operation of subtracting 1, then maybe I can simplify the expression.

But regardless of the method, all paths lead to the answer 7.

Therefore, I can be confident that the value of g(f(5) - 1) is 7.

**Final Answer**
The value of $ g(f(5) - 1) $ is \boxed{7}.

To solve the expression $ g(f(5) - 1) $, we proceed step by step, evaluating the inner function first and then applying the outer function.

---

### Step 1: Evaluate $ f(5) $

We are given the function:

$$
f(x) = 2x - 3
$$

Substitute $ x = 5 $:

$$
f(5) = 2(5) - 3 = 10 - 3 = 7
$$

---

### Step 2: Compute $ f(5) - 1 $

Now subtract 1 from the result:

$$
f(5) - 1 = 7 - 1 = 6
$$

---

### Step 3: Evaluate $ g(6) $

We are given the function:

$$
g(x) = x + 1
$$

Substitute $ x = 6 $:

$$
g(6) = 6 + 1 = 7
$$

---

### Final Answer

Thus, the value of $ g(f(5) - 1) $ is:

$$
\boxed{7}
$$

[Assistant]
Explanation:
Okay, let me try to figure out this problem. So, we have two functions, f(x) = 2x - 3 and g(x) = x + 1. The question is asking for the value of g(f(5) - 1). Hmm, let me break this down step by step.

First, I need to find f(5). Since f(x) is defined as 2x - 3, substituting x with 5 should give me the value of f(5). Let me write that out:

f(5) = 2*(5) - 3.

Calculating that, 2 times 5 is 10, and then subtracting 3 gives me 10 - 3, which equals 7. So, f(5) is 7.

Now, the next part of the problem is g(f(5) - 1). Since I already found that f(5) is 7, I can substitute that into the expression. So, inside the parentheses, it's 7 - 1. Let me compute that: 7 minus 1 is 6.

So now, the problem simplifies to finding g(6). Because g(f(5) - 1) becomes g(6).

Now, let me recall what g(x) is. It's given as x + 1. Therefore, substituting x with 6, we have:

g(6) = 6 + 1.

Adding those together, 6 + 1 equals 7.

Wait, so putting it all together, g(f(5) - 1) equals 7? Let me check my steps again to make sure I didn't make a mistake.

First, f(5): 2*5 is 10, minus 3 is 7. That seems right. Then subtract 1 from that result: 7 - 1 is 6. Then plug into g(x): 6 + 1 is 7. Yeah, that seems correct.

But let me think again. Maybe I misread the problem? Let me check the original question again. It says g(f(5) - 1). So, is it g of (f(5) minus 1), or is there a different order of operations? For example, could it be interpreted as g(f(5 - 1))? But no, the parentheses are around f(5) - 1, so it's definitely g evaluated at (f(5) - 1).

Alternatively, maybe I need to consider if there's a different way to parse the expression? Let me confirm. The expression is g(f(5) - 1). Since function composition is usually written with parentheses, but here it's just a function g applied to the result of f(5) minus 1. So, yes, first compute f(5), subtract 1, then apply g.

So, my steps are correct. Let me verify once more:

f(5) = 2*5 - 3 = 10 - 3 = 7.

Then, f(5) - 1 = 7 - 1 = 6.

Then, g(6) = 6 + 1 = 7.

Therefore, the value of g(f(5) - 1) is 7.

Wait, but just to be thorough, let me check if there's another way someone might approach this problem and maybe make a mistake. For example, someone might think that g(f(5) - 1) is the same as g(f(5 - 1)), which would be different. Let me see what that would be. If someone incorrectly interprets the parentheses, they might compute f(5 - 1) first. Let's see:

If it were g(f(5 - 1)), then 5 - 1 is 4, then f(4) is 2*4 - 3 = 8 - 3 = 5, then g(5) is 5 + 1 = 6. But that's different from 7. However, the original problem is g(f(5) - 1), not g(f(5 - 1)). So, the parentheses are important here.

Therefore, as long as we follow the order of operations and the parentheses, we should be okay.

Another possible mistake could be miscalculating f(5). Let me check that again. f(x) = 2x - 3. So, plugging in 5: 2*5 is 10, minus 3 is 7. Correct.

Then subtracting 1: 7 - 1 is 6. Then g(6) is 6 + 1, which is 7.

Alternatively, maybe someone might think that g(f(5) - 1) is g(f(5)) - 1? Let me check that. If that were the case, then g(f(5)) would be g(7) which is 7 + 1 = 8, then subtract 1 would be 7. Wait, so in this case, both interpretations give the same result?

Wait, that's interesting. Let me think. If you have g(f(5) - 1), is that the same as g(f(5)) - 1?

No, not necessarily. Because g is a function, so g(a - b) is not the same as g(a) - b unless the function has specific properties.

But in this case, let me check. If we compute g(f(5) - 1) as per the original problem, we get 7. If we compute g(f(5)) - 1, that would be (g(7)) - 1 = 8 - 1 = 7. So, in this case, both ways give the same answer.

But that's a coincidence? Let me check with different numbers. Suppose f(5) was 8 instead of 7. Then, g(f(5) - 1) would be g(7) = 8. Whereas g(f(5)) - 1 would be g(8) - 1 = 9 - 1 = 8. So, same result. Wait, so is this always true?

Wait, let me think. If g(x) = x + 1, then g(a - 1) = (a - 1) + 1 = a. And g(a) - 1 = (a + 1) - 1 = a. So, in this case, they are equal.

Therefore, in this specific problem, because g(x) is x + 1, then g(f(5) - 1) equals g(f(5)) - 1. So, even if someone misinterprets the parentheses, they might still get the same answer. But that's because of the specific form of g(x).

But in general, this is not true. For example, if g(x) was something else, like g(x) = 2x, then g(a - 1) would be 2(a - 1) = 2a - 2, whereas g(a) - 1 would be 2a - 1. Which are different.

So, in this problem, because of the specific function g(x) = x + 1, both interpretations give the same result. But the correct way is to follow the parentheses.

But since both ways give the same answer here, maybe the problem is designed that way? But regardless, the correct approach is to compute f(5) first, subtract 1, then apply g.

So, in conclusion, the answer is 7.

But just to make sure, let me write all steps again:

1. Compute f(5):

f(5) = 2*(5) - 3 = 10 - 3 = 7.

2. Compute f(5) - 1:

7 - 1 = 6.

3. Compute g(6):

g(6) = 6 + 1 = 7.

Therefore, the value of g(f(5) - 1) is 7.

Yes, that seems consistent.

Alternatively, if I wanted to write this as a composition of functions, maybe? Let me think. But the problem isn't asking for a composition, it's asking for g evaluated at (f(5) - 1). So, no, it's not a composition of functions, it's a function evaluated at an expression.

Alternatively, if it were g(f(x)) - 1, then we would have a different story, but no, the problem is specifically g(f(5) - 1).

Another way to think about it: maybe substitute the functions into each other? Let me try that.

First, let's find f(5) - 1. As we found, that's 7 - 1 = 6. Then, g(6) is 6 + 1 = 7.

Alternatively, if I wanted to write an expression for g(f(x) - 1), I could do that. Let me try:

g(f(x) - 1) = [f(x) - 1] + 1 = f(x) - 1 + 1 = f(x).

Wait, that's interesting. So, if you have g(f(x) - 1), since g is x + 1, then substituting f(x) - 1 into g gives [f(x) - 1] + 1 = f(x). Therefore, in general, g(f(x) - 1) = f(x).

Therefore, if we plug in x = 5, then g(f(5) - 1) = f(5) = 7. Which matches our previous result.

So, that's another way to see it. Since g is the function that adds 1, subtracting 1 before applying g cancels out the addition. Therefore, g(f(x) - 1) simplifies to f(x).

Therefore, regardless of the value of x, g(f(x) - 1) is equal to f(x). Therefore, for x = 5, it's f(5) = 7.

This seems like a more general approach. So, if I can recognize that g is the inverse operation of subtracting 1, then maybe I can simplify the expression.

But regardless of the method, all paths lead to the answer 7.

Therefore, I can be confident that the value of g(f(5) - 1) is 7.

**Final Answer**
The value of $ g(f(5) - 1) $ is \boxed{7}.

To solve the expression $ g(f(5) - 1) $, we proceed step by step, evaluating the inner function first and then applying the outer function.

---

### Step 1: Evaluate $ f(5) $

We are given the function:

$$
f(x) = 2x - 3
$$

Substitute $ x = 5 $:

$$
f(5) = 2(5) - 3 = 10 - 3 = 7
$$

---

### Step 2: Compute $ f(5) - 1 $

Now subtract 1 from the result:

$$
f(5) - 1 = 7 - 1 = 6
$$

---

### Step 3: Evaluate $ g(6) $

We are given the function:

$$
g(x) = x + 1
$$

Substitute $ x = 6 $:

$$
g(6) = 6 + 1 = 7
$$

---

### Final Answer

Thus, the value of $ g(f(5) - 1) $ is:

$$
\boxed{7}
$$

Solution: We have $f(5) = 2(5) -3 = 7$, so $g(f(5)-1) = g(7-1) = g(6) = 6+1 = \boxed{7}$.