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From the following equalities, I'm supposed to find a law and prove it by induction.

$1=1$; $1-4=-(1+2)$; $1-4+9=1+2+3$, and $1-4+9-16=-(1+2+3+4)$.

I supposed the law was $\sum^{n}_{k=1} (-1)^{k+1}k^2=(-1)^{n+1}\sum^n_{k=1} k$.

Let's say it's valid for a $n\in \mathbb{N}$. For $n+1$, I have $\sum^{n+1}_{k=1} (-1)^{k+1}k^2=\sum^{n}_{k=1} (-1)^{k+1}k^2+(-1)^{n+2}(n+1)^2$, which by the induction hypothesis it's equal to $(-1)^{n+1}\sum^n_{k=1} k+(-1)^{n+2}(n+1)^2$.

I'm stuck here. What's the 'trick'?

Any help would be appreciated.

David Holden
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An old man in the sea.
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3 Answers3

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although this is not an inductive proof it is worth noting, in passing, that this rather pretty result follows from the fact that the difference of the squares of two consecutive integers is equal to their sum. thus, for example:

$$ 9^2-8^2 = 9+8\\ 7^2-6^2 = 7+6\\ 5^2-4^2 = 5+4\\ 3^2-2^2= 3+2\\ 1^2-0^2= 1+0 $$ adding these equations gives $$ 9^2-8^2+7^2-6^2+5^2-4^2+3^2-2^2+1^2 = 9+8+7+6+5+4+3+2+1 $$

David Holden
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You have to prove that $$(-1)^{n+1}\sum^n_{k=1} k+(-1)^{n+2}(n+1)^2=(-1)^{n+2}\sum^{n+1}_{k=1} k$$ that is $$(n+1)^2=\sum^{n+1}_{k=1} k+\sum^n_{k=1} k=2\sum^n_{k=1} k+(n+1)$$ or $$\sum^n_{k=1} k=\frac{n(n+1)}{2}$$ which holds (see Proof for formula for sum of sequence $1+2+3+\ldots+n$?).

Community
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Robert Z
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You need to show $$(-1)^{n+1}\sum_{k=1}^nk+(-1)^{n+2}(n+1)^2=(-1)^{n+2}\sum_{k=1}^{n+1}k.$$ Dividing by $(-1)^{n+1}$, this is equivalent to $$-\sum_{k=1}^nk+(n+1)^2=\sum_{k=1}^{n+1}k.$$ Now you can just apply the formula for $1+2+\dots+n$.

pi66
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