<s>[INST] Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer 
Prove that $1^2 - 2^2 + 3^2 - 4^2 + \cdots + (-1)^{n-1}n^2 = \frac12(-1)^{n-1} n (n + 1)$, where $n $ is a positive integer

How do I prove the above expression using mathematical induction? So far I have only been proving simpler stuff. The base case P(1) is easy enough but I am lost as to where I should even start with my inductive step. I really don't know what the steps for P(k + 1) should be, and so help would be greatly appreciated.
 [/INST] We are to prove 
$$\sum_{k=1}^n(-1)^{k-1}k^2=\frac{(-1)^{n-1}n(n+1)}{2} \tag 1$$
First, we establish a base case.  So, for $n=2$, we note that the left-hand side of $(1)$ is $1^2-2^2=-3$ while the right-hand side of $(1)$ is $(-1)\frac{2\times 3}{2}=-3$.  Thus, the case $n=2$ is verified.
Next, we assume that $(1)$ holds for some number $N$ and demonstrate that it holds for $N+1$.  
To that end, we have 
$$\begin{align}
\sum_{k=1}^{N+1}(-1)^{k-1}k^2&=\sum_{k=1}^{N}(-1)^{k-1}k^2+(-1)^N(N+1)^2\\\\
&=\frac{(-1)^{N-1}N(N+1)}{2} +(-1)^N(N+1)^2\\\\
&=\frac{(-1)^{N+1}}{2}\left(N(N+1)-2(N+1)^2\right)\\\\
&=\frac{(-1)^N(N+1)(N+2)}{2}
\end{align}$$
and this completes the proof by induction.
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