<s>[INST] Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$. Use induction to prove that sum of the first $2n$ terms of the series $1^2-3^2+5^2-7^2+…$ is $-8n^2$.
I came up with the formula $\displaystyle\sum_{r=1}^{2n} (-1)^{r+1}(2r-1)^2=-8n^2$ but I got stuck proving it by induction. Should I use another formula or is it correct and I should continue trying to prove it by induction?
 [/INST] First, show that this is true for $n=1$:
$\sum\limits_{r=1}^{2}(-1)^{r+1}\cdot(2r-1)^2=-8$
Second, assume that this is true for $n$:
$\sum\limits_{r=1}^{2n}(-1)^{r+1}\cdot(2r-1)^2=-8n^2$
Third, prove that this is true for $n+1$:
$\sum\limits_{r=1}^{2(n+1)}(-1)^{r+1}\cdot(2r-1)^2=$
$\color{red}{\sum\limits_{r=1}^{2n}(-1)^{r+1}\cdot(2r-1)^2}+\color{green}{(-1)^{(2n+1)+1}\cdot(2(2n+1)-1)^2}+\color{blue}{(-1)^{(2n+2)+1}\cdot(2(2n+2)-1)^2}=$
$\color{red}{-8n^2}+\color{green}{(-1)^{(2n+1)+1}\cdot(2(2n+1)-1)^2}+\color{blue}{(-1)^{(2n+2)+1}\cdot(2(2n+2)-1)^2}=$
$-8n^2-8(2n+1)=$
$-8(n^2+2n+1)=$
$-8(n+1)^2$

Please note that the assumption is used only in the part marked red.
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