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

Question

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.

Answer 1

$$P(k)=\frac{(-1)^{k-1} \cdot k \cdot (k + 1)}{2}$$ 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.

$$1^2 - 2^2 + 3^2 - 4^2 + ... + (-1)^{k-1}k^2+(-1)^k(k+1)^2 = P(k)+(-1)^k(k+1)^2\quad =\frac{(-1)^{k-1} \cdot k \cdot (k + 1)}{2} + (-1)^k(k+1)^2 =\frac{(-1)^{k-1}k\cdot (k+1) + (-1)^k2\cdot(k+1)^2}{2}\quad = \frac{(-1)^{k}(-k\cdot (k+1) + 2\cdot(k+1)^2)}{2}\quad = \frac{(-1)^{k}(-k\cdot (k+1) + 2\cdot(k+1)^2)}{2} \quad = \frac{(-1)^k(k^2+3k+2)}{2}=P(k+1)$$

Answer 2

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.

Answer 3

If $n$ is even, $$ \begin{align} \sum_{k=1}^n(-1)^{k-1}k^2 &=\sum_{k=1}^{n/2}\left[(2k-1)^2-(2k)^2\right]\\ &=\sum_{k=1}^{n/2}\left[1-4k\right]\\ &=-\frac{n^2+n}2 \end{align} $$ If $n$ is odd, $n-1$ is even. $$ \begin{align} \sum_{k=1}^n(-1)^{k-1}k^2 &=n^2+\sum_{k=1}^{n-1}(-1)^{k-1}k^2\\ &=n^2-\frac{n^2-n}2\\ &=\frac{n^2+n}2 \end{align} $$ Therefore, $$ \sum_{k=1}^n(-1)^{k-1}k^2=(-1)^{n+1}\frac{n^2+n}2 $$