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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.

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    $\begingroup$ The sequence on the LHS is not very clear. $\endgroup$
    – parsiad
    Sep 13, 2015 at 5:18
  • $\begingroup$ You need something squared as the last term on the left side? $\endgroup$
    – 2'5 9'2
    Sep 13, 2015 at 5:35
  • $\begingroup$ If the last term on the left hand side is not $(-1)^{n-1}n^2$, then it is not clear what the general term of the sum is supposed to be. $\endgroup$
    – robjohn
    Sep 13, 2015 at 5:45
  • $\begingroup$ I don't know what I need to add (if I should add anything)to the LHS so that RHS will eventually become $$\frac{(-1)^n (n + 1)(n + 2)}{2}$$ $\endgroup$ Sep 13, 2015 at 5:46
  • $\begingroup$ I'm sorry for your troubles, but I merely copied the expression that I found in the material, and by now, seeing the expression being under the "Proof by Induction" topic confuses me. $\endgroup$ Sep 13, 2015 at 5:49

3 Answers 3

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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, $$ \sum_{k=1}^n(-1)^{k-1}k^2=(-1)^{n+1}\frac{n^2+n}2 $$

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  • $\begingroup$ This is, of course, the way to approach this problem. While, the OP seems to be requesting a proof by induction, it's always good to show multiple ways of attack! $\endgroup$
    – Mark Viola
    Sep 13, 2015 at 5:40
  • $\begingroup$ The question was not originally stated in a fashion that I could tell what the right hand side was supposed to be. It is hard to proceed with induction without knowing the final result, so what I did was to compute the right hand side. Now that the right hand side is readable, I could delete this answer and then the proofs by induction can stay. However, the OP has changed the question to something which makes the left hand side confusing, $\endgroup$
    – robjohn
    Sep 13, 2015 at 5:43
  • $\begingroup$ I took a guess at the intended problem, read "How do I prove the above expression using mathematical induction?" in the OP, and proceeded accordingly. Perhaps I read a later version with more info. Anyway, I always enjoy reading your solutions. $\endgroup$
    – Mark Viola
    Sep 13, 2015 at 5:57
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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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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. $\endgroup$
    – Mark Viola
    Sep 13, 2015 at 13:59
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$$P(k)=\frac{(-1)^{k-1} \cdot k \cdot (k + 1)}{2}$$ 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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  • $\begingroup$ I know that $P(k + 1)$ should look like this, but I don't know how to get there from just $P(k)$ in the inductive step. Thank you anyway, good sir. $\endgroup$ Sep 13, 2015 at 5:29
  • $\begingroup$ You mean how to get the relation between $P(k)$ and $P(k+1)$ in a formular? $\endgroup$
    – Antitheos
    Sep 13, 2015 at 5:30
  • $\begingroup$ Yes. I don't know how to get to the expression $P(k + 1)$ when I'm doing the actual proof, so that I may show that $P(k + 1)$ follows $P(k)$. $\endgroup$ Sep 13, 2015 at 5:33
  • $\begingroup$ I'll add it. (Changed your wording in the question to make it clear.) $\endgroup$
    – Antitheos
    Sep 13, 2015 at 5:40
  • $\begingroup$ In the penultimate step, you should have $\frac{(-1)^k(k^2 + 3k + 2)}{2}$. You can verify this by expanding $\frac{(-1)^k[-k(k + 1) + 2(k + 1)^2]}{2}$. Note that $k^2 + 3k + 2 = (k + 1)(k + 2)$, while $k^2 + k + 2$ is irreducible with respect to the rationals. $\endgroup$ Sep 14, 2015 at 10:06

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