Find the expansion of each of the following generating functions:?

Start with 1/(1-x) = sum(n=0 to infinity) x^n.

(a) Differentiating yields
1/(1 - x)^2 = sum(n=1 to infinity) nx^(n-1).

Replace x with (-2x):
1/(1 + 2x)^2 = sum(n=1 to infinity) n(-2)^(n-1) x^(n-1).
[ = sum(n=0 to infinity) (n+1)(-2)^n x^n.]

Multiply both sides by x:
x/(1 + 2x)^2 = sum(n=1 to infinity) n(-2)^(n-1) x^n.

Thus,
(x - 1)/(1 + 2x)^2
= -1 + sum(n=1 to infinity) [n(-2)^(n-1) - (n+1)(-2)^n] x^n
= -1 + sum(n=1 to infinity) [n(-2)^(n-1) + (2n+2)(-2)^(n-1)] x^n
= -1 + sum(n=1 to infinity) (3n+2)*(-2)^(n-1) x^n
= sum(n=0 to infinity) (3n+2)*(-2)^(n-1) x^n.

(b) Replace x with x^2 in the Geometric Series:
1/(1 - x^2) = sum(n=0 to infinity) x^(2n).

Multiply both sides by x:
x/(1 - x^2) = sum(n=0 to infinity) x^(2n+1).

(c) Note that (1 - 5x + 6x^2) = (1 - 2x)(1 - 3x).

By Partial Fractions,
(1 - 7x)/(1 - 5x + 6x^2) = A/(1 - 2x) + B/(1 - 3x).

1 - 7x = A(1 - 3x) + B(1 - 2x).

x=1/2: -5/2 = -A/2 ==> A = 5
x =1/3: -4/3 = B/3 ==> B = -4

Thus,
(1 - 7x)/(1 - 5x + 6x^2)
= 5/(1 - 2x) - 4/(1 - 3x)
= 5 * sum(n=0 to infinity) (2x)^n - 4 * sum(n=0 to infinity) (3x)^n, via geometric series
= sum(n=0 to infinity) (5 * 2^n - 4 * 3^n) x^n.

I hope this helps!