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$$\sum_{i=1}^n (-1)^{i-1} i^2 = (-1)^{n-1} \sum_{i=1}^n i$$ for all integers $n \ge 1$.

Gerry Myerson
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Anh Luu
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    Welcome to MSE! We condemn uploading pictures, you should try to write your problem using LaTeX. Here is your LaTeX guide: https://oeis.org/wiki/List_of_LaTeX_mathematical_symbols Also please write whatever you have tried to solve the problem. – Anik Bhowmick Sep 05 '22 at 02:51
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    This seems like a good place to use induction. – angryavian Sep 05 '22 at 02:55
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  • I would suggest trying a few to see if you get a feel and maybe even an intuition as to why it would be: $n-1: 1 = 1$, $n=2: 1-4=-(1+2)$, $n=3: 1-4+9 = (1+2+3)$ and $n=4: 1-4+9-16 = -(1+2+3+4)$. Are you seeing any patterns? Any reason to see why it may be true? It seems like when you have $n^2$ and you subtract $\sum_{i=1}^{n-1} i$ you will end up with $\sum_{i=1}^n i$. Why do you think that would be? – fleablood Sep 05 '22 at 03:33

1 Answers1

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Let $n=2m$, then $$\sum_{1}^{2m} (-1)^{k-1}*k^2=-\sum_{k=1}^m (4k-1)=-4\frac{m(m+1)}{2}-m=-m(2m+1)=(-1)^{n-1}\frac{n(n+1)}{2}=(-1)^{n-1}\sum_{k=1}^{n} k.$$ Next, take $n=2m+1$. then

$$S=\sum_{k=1}^{2m+1}(-1)^{k-1} k^2=\sum_{k=1}^{2m}(-1)^{k-1} k^2+(2m+1)^2=-m(2m+1)+4m^2+4m+1=2m^2+3m+1=(2m+1)(m+1)=\frac{(2m+1)(2m+2)}{2}=(-1)^{n-1}\frac{n(n+1)}{2}=(-1)^{n-1}\sum_{k=1}^{n} k$$

Z Ahmed
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