Arithmetic-geometric progression;Sn? $S_n=1+ 3x^2+ 5x^4 +7x^6 +...+(2n-1)x^{(2n-2)}$
a)Write down the series for $x^2S_n$. 
Hence show that $S_n=\frac{1}{(1-x^2)} + (2x^2)\frac{(1-x^{(2n-2)})}{(1-x^2)}- (2n-1)\frac{(x^{2n})}{(1-x^2)}$, where $x$ cannot be $\pm1$
The series for $x^2S_n = x^2 +3x^4 + 5x^6 +7x^8+...     
+(2n-3)(x^{(2n)})$
So how to show?
 A: So here is the way to go about it:
$$S_n = 1 + 3x^2 + 5x^4 + \cdots + (2n-1)x^{(2n-2)} \rightarrow$$
$$x^2S_n = x^2 + 3x^4 + 5x^6 + \cdots + (2n-1)x^{2n}$$
Now if we subtract $x^2S_n$ from $S_n$ term by term we get:
$$S_n - x^2S_n = 1 + (3x^2 - x^2) + (5x^4 - 3x^4) + \cdots + [(2n-1)x^{2n-2} - (2n-3)x^{2n-2}] - (2n-1)x^{2n} =$$
$$= 1 + 2x^2 + 2x^4 + \cdots + 2x^{2n-2} - (2n-1)x^{2n}$$
Now, on the left hand side we factor $S_n$ out to get $(1-x^2)S_n$. On the right hand side we group the terms in 3 parts:
$$(1-x^2)S_n = [1] + [2x^2 + 2x^4 + \cdots + 2x^{2n-2}] - [(2n-1)x^{2n}]$$
Now the middle group is a geometric series:
$$G_n = 2x^2 + 2x^4 + \cdots + 2x^{2n-2} = 2(x^2 + x^4 + \cdots + x^{2n-2}) = 2\times \frac{x^2 - x^{2n}}{(1 - x^2)} = 2x^2\times \frac{1 - x^{2n-2}}{(1 - x^2)}$$
Plugging in we have $$(1-x^2)S_n = [1] + [2x^2\times \frac{(1 - x^{2n-2})}{(1 - x^2)}] - [(2n-1)x^{2n}]$$
Now dividing both sides by $(1 - x^2)$ you get:
$$S_n = \frac{1}{(1 - x^2)} + \frac{2x^2(1-x^{2n-2})}{(1-x^2)^2} - \frac{(2n-1)x^{2n}}{(1-x^2)}$$
