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I'm trying to prove following statement through induction:

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

I have only seen how to prove with induction when the variable $n$ is not included in the sum function. Like here (this statement is not correct, just to provide an example):

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

It confuses me and I don't know how to proceed. Any advice?

Thank you.

Martin Sleziak
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Stanko
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  • @yedaynara Are you basing on the second statement? Because that's just an example and an incorrect statement. – Stanko Jan 10 '15 at 14:47
  • @yedaynara My fault, I did not clarify that, I made an edit now. – Stanko Jan 10 '15 at 14:49
  • Yes, I should have checked that before commenting. My fault too! BTW I deleted my comment as it is no use keeping incorrect answers. – Landon Carter Jan 10 '15 at 14:50
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    Well, $\sum(-1)^kk^2=-1+2^2-3^2+4^2-...+(-1)^nn^2$ Can you try to use the form $a^2-b^2=(a+b)(a-b)$ here? You will get the sum of first $n$ positive integers, or some truncated version of that, along with some sign. The case $(-1)^{n-k}$ should be dealt in the same way as I had mentioned in my "incorrect" comment. – Landon Carter Jan 10 '15 at 14:52

3 Answers3

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

kingW3
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We can derive from scratch as well. For even $n=2 m$, we can simply pair up the summands to get

$$\sum_{k=1}^{m} ((2 k)^2 - (2 k-1)^2) = \sum_{k=1}^{m} (4 k-1) = 2 m (m+1) - m = \frac{n (n+1)}{2} $$

For odd $n=2 m+1$, we may do the same, but we have a last (positive) term, and we change the sign of the even part of the sum. Thus we get

$$(2 m+1)^2 - 2 m (m + 1) + m = 2 m^2 + 3 m + 1 = 2 \left (\frac{n-1}{2} \right )^2 + 3 \left (\frac{n-1}{2} \right ) + 1 = \frac{n (n+1)}{2}$$

Ron Gordon
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By induction: for $n=1$: $$\sum_{k=1}^{1} (-1)^{1-k}k^2 = (-1)^0\cdot 1^2=1=\frac{1\cdot (1+1)}{2} \qquad \checkmark$$ Assume it is true for $n=m$: $$\sum_{k=1}^{m} (-1)^{m-k}k^2 = \frac{m\cdot (m+1)}{2}$$ Prove for $n=m+1$: $$\begin{align}\sum_{k=1}^{m+1} (-1)^{m+1-k}k^2 = &\sum_{k=1}^{m} (-1)^{m+1-k}k^2 + (-1)^{m+1-(m+1)}(m+1)^2=\\ &(-1)\sum_{k=1}^m(-1)^{m-k}k^2+(m+1)^2=\\ &(-1)\cdot \frac{m\cdot(m+1)}{2}+(m+1)^2=\\ &(m+1)(m+1-\frac m2)=\\ &\frac{(m+1)\cdot (m+2)}{2}. \qquad \checkmark \end{align}$$ Note: There is a deleted answer (visible for users with $10,000+$ rep), where there was a sign mistake.

farruhota
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