Problem source

\begin{questionparts} 
\item By considering $(1+x+x^{2}+\cdots+x^{n})(1-x)$ show that, if $x\neq1$, 
\[
1+x+x^{2}+\cdots+x^{n}=\frac{1-x^{n+1}}{1-x}.
\]
\item By differentiating both sides and setting $x=-1$ show that
\[
1-2+3-4+\cdots+(-1)^{n-1}n
\]
takes the value $-n/2$ is $n$ is even and the value $(n+1)/2$ if
$n$ is odd.
\item Show that 
\[
1^{2}-2^{2}+3^{2}-4^{2}+\cdots+(-1)^{n-1}n^{2}=(-1)^{n-1}(An^{2}+Bn)
\]
where the constants $A$ and $B$ are to be determined. 
\end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
&& (1+x+x^{2}+\cdots+x^{n})(1-x) &= 1-x+x-x^2+\cdots -x^n+x^n-x^{n+1} \\
&&&= 1-x^{n+1} \\
\Rightarrow && 1+x+x^2+\cdots+x^n &= \frac{1-x^{n+1}}{1-x} \tag{dividing by $1-x$}
\end{align*}
\item \begin{align*}
\frac{\d}{\d x}: && 0+1+2x+\cdots+nx^{n-1} &= \frac{(n+1)(1-x)x^n+(1-x^{n+1})}{(1-x)^2} \\
\Rightarrow && 1-2x+\cdots+(-1)^n n &= \frac{-(n+1)2(-1)^n+(1-(-1)^{n+1})}{4} \\
&&&= \begin{cases} \frac{-(n+1)\cdot2\cdot1+(1-(-1)}{4} & \text{if }n\text{ is even} \\
\frac{-(n+1)\cdot 2 \cdot(-1)+(1-1)}{4} & \text{if }n\text{ is odd}\end{cases} \\
&&&= \begin{cases} \frac{-n}{2} & \text{if }n\text{ is even}\\
\frac{n+1}{2} & \text{if }n\text{ is odd}\end{cases} \\
\end{align*}
\item \begin{align*}
x: && x+2x^2+\cdots+nx^{n} &= \frac{(n+1)(1-x)x^{n+1}+x(1-x^{n+1})}{(1-x)^2} \\
&&&= \frac{x+(n+1)x^{n+1}-nx^{n+2}}{(1-x)^2}\\
\frac{\d}{\d x}: && 1^2+2^2x + \cdots + n^2x^{n-1} &= \frac{(1-x)^2(1+(n+1)^2x^{n}-n(n+2)x^{n+1}) +2(1-x)(x+(n+1)x^{n+1}-nx^{n+2})}{(1-x)^4} \\
&&&= \frac{1 + x - (1 + n)^2 x^n + (2 n^2+2n-1) x^{n+1} - n^2 x^{n+2}}{(1-x)^3} \\
\Rightarrow && 1^2-2^2 + \cdots + (-1)^{n-1}n^2 &= \frac{(-1)^n \l - (1 + n)^2- (2 n^2+2n-1) - n^2 \r}{8} \\
&&&= \frac{(-1)^n(-4n^2-4n}{8} \\
&&&= \frac{(-1)^{n-1}(n^2+n)}{2}
\end{align*}
\end{questionparts}