<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] $$P(k)=\frac{(-1)^{k-1} \cdot k \cdot (k + 1)}{2}$$
Therefore
$$P(k+1)=\frac{(-1)^{k} \cdot (k+1) \cdot (k + 2)}{2}$$
The way to go: 
Write $P(k+1)$ as something containing $P(k)$.
Since you seem are asking only about some starting help, this should suffice. =)
After reading your comment:
The general way to go is starting from both direction and either go all the way are meet in the middle.
$$1^2 - 2^2 + 3^2 - 4^2 + ... + (-1)^{k-1}k^2+(-1)^k(k+1)^2 = P(k)+(-1)^k(k+1)^2\quad =\frac{(-1)^{k-1} \cdot k \cdot (k + 1)}{2} + (-1)^k(k+1)^2 =\frac{(-1)^{k-1}k\cdot (k+1) + (-1)^k2\cdot(k+1)^2}{2}\quad = \frac{(-1)^{k}(-k\cdot (k+1) + 2\cdot(k+1)^2)}{2}\quad = \frac{(-1)^{k}(-k\cdot (k+1) + 2\cdot(k+1)^2)}{2} \quad = \frac{(-1)^k(k^2+3k+2)}{2}=P(k+1)$$ 
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