Find sum of series $S = 1 + 3x + 5x^2 + ... + (2n + 1)x^n$, using $(1-x)S$ I study maths as a hobby. I am stuck on the following problem, which is an A level question from the Oxford board set probably in the '70s:
If $S = 1 + 3x + 5x^2 + ... + (2n + 1)x^n$, prove, by considering $(1 - x)S$ or otherwise, that if $x \ne 1$,
$$S = \frac{1 + x - (2n + 3)x^{n + 1} + (2n + 1)x^{n + 2}}{(1 - x^2)}$$
Firstly, I am puzzled that the nth term of S is given as $(2n + 1)x^n$.
For example, its 3rd term, $5x^2$ should surely be $7x^3$. Or does the numbering of n start at $n = 0$?
However,
$S =  1 + 3x + 5x^2 + ... + (2n + 1)x^n$
$Sx = x + 3x^2 + 5x^3 + ... + (2n + 1)x^{n + 1}$
$(1 - x)S = 1 + 2x + 2x^2 + ...+ (2n +1)x^{n+1}$
But I don't see how to get from this to the needed proof.
 A: The numbering starts at $n = 0$.
Your evaluation is not quite complete.  Instead,
$$\begin{align}
S &= 1 + 3x + 5x^2 + 7x^3 + \cdots + \hphantom{(\vphantom{} - 1)}(2n + 1)x^n \\
xS &= \hphantom{1 + 3}x + 3x^2 + 5x^3 + \cdots + (2(n-1)+1) x^n + (2n + 1)x^{n+1} \\
(1-x)S &= 1 + 2x + 2x^2 + 2x^3 + \cdots + \hphantom{((n-1)+1)}2x^n \color{red}{- (2n+1)x^{n+1}}.
\end{align}$$
Therefore, $$(1-x)S + 1 + (2n+1)x^{n+1} = 2 + 2x + 2x^3 + \cdots + 2x^n = 2(1 + x + x^2 + \cdots + x^n).$$  Since we know that the sum of a finite geometric series with common ratio $x$ is $$1 + x + x^2 + \cdots + x^n = \frac{x^{n+1} - 1}{x - 1},$$ it follows that $$(1-x)S = \frac{2(x^{n+1} - 1)}{x-1} - 1 - (2n+1)x^{n+1},$$ and dividing by $1-x$ and simplifying gives the desired value of $S$.

For the OP's benefit:
$$\begin{align}
S &= \frac{1}{1-x} \left( \frac{2(x^{n+1} - 1)}{x-1} - 1 - (2n+1)x^{n+1} \right) \\
&= -\frac{2(x^{n+1} - 1) - (x-1) - (2n+1)x^{n+1}(x-1)}{(x-1)^2} \\
&= -\frac{2x^{n+1} - 2 - x + 1 - (2n+1) x^{n+2} + (2n+1) x^{n+1}}{(x-1)^2} \\
&= \frac{(2n+1) x^{n+2} - (2n+3)x^{n+1} + x + 1}{(x-1)^2}.\end{align}$$
