Final answer:
The closed form of the generating function for the arithmetic progression (a, a + d, a + 2d, a + 3d, ...) is a/(1-d). Plugging in x = 1, the generating function evaluated at x = 1 gives us the sum of the arithmetic progression, which simplifies to a + d.
Step-by-step explanation:
The closed form of the generating function for the arithmetic progression (a, a + d, a + 2d, a + 3d, ...) is given by a/(1-d).
To understand why, let's consider the generating function of the arithmetic progression. Let's call it G(x). Each term in the arithmetic progression can be represented as a power of x multiplied by the corresponding term (a, a + d, a + 2d, a + 3d, ...). So, G(x) can be written as:
G(x) = a + (a + d)x + (a + 2d)x^2 + (a + 3d)x^3 + ...
We can rewrite this as:
G(x) = a + ax + dx^1 + ax^2 + 2dx^2 + ax^3 + 3dx^3 + ...
Now, we can factor out the terms with the same power of x:
G(x) = a(1 + x + x^2 + x^3 + ...) + d(x + 2x^2 + 3x^3 + ...)
Using the formula for the sum of an infinite geometric series, we can simplify this expression to:
G(x) = a/(1 - x) + dx/(1 - x)^2
So, the closed form of the generating function is a/(1 - x) + dx/(1 - x)^2. In the given arithmetic progression, x represents the common ratio (d) and we are interested in the value of the generating function when x = 1. Plugging in x = 1, we get:
G(1) = a/(1 - 1) + d/(1 - 1)^2 = a + d
Therefore, the generating function evaluated at x = 1 gives us the sum of the arithmetic progression. In this case, it simplifies to a + d.