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I would like to prove the following equation using induction. However that seems somehow impossible at least for me:

$\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$

I tried to show that some property $E()$ holds for $n=1$, $E(1)$ but I just get something which doesnt makes sense:

$(-1)^1 \cdot 1^2=(2 \cdot 1+1) \cdot 1$

$-1=3$

Mamba
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  • Minor quibble: you don't solve an equation using induction, you prove an equality. – mrp Mar 26 '15 at 10:09
  • If you try for $n=0$, instead, the summation at the left hand side is empty, so it means $0$; the right hand side is $0$ as well. – egreg Mar 26 '15 at 10:37
  • I thought because of $k=1$ I have to begin with $n=1$ – Mamba Mar 26 '15 at 10:40

1 Answers1

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When $n=1$, the sum on the left hand side is from $k=1$ to $k=2$, which is

$$(-1)^1 \cdot 1^2 + (-1)^2 \cdot 2^2 = (-1) + 4 = 3$$

Empiricist
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