<s>[INST] Proving that the sum of the first $2n$ terms of the series $1^2 - 3^2 + 5^2 - \cdots$ is $-8n^2$ by induction 
Use mathematical induction to prove the following for the first $2n$ terms of the series $$1^2 - 3^2 + 5^2 - 7^2 + \cdots = -8n^2.$$

As we have odd numbers that are squared we could use $n = 2k-1$.
But the $2$ sides do not equate for $n=1$ or $2k-1$ (if you set $k=1$).
Also need to find the sum to $2n+1$ terms.
 [/INST] $1^2−3^2+5^2−7^2+⋯=−8n^2$
Last term of this series is $(4n-3)^2-(4n-1)^2$
$\Rightarrow P(n):1^2−3^2+5^2−7^2+⋯+(4n-3)^2-(4n-1)^2=−8n^2$
For n=1
$\Rightarrow P(1):1^2−3^2=−8(1)^2$
$\Rightarrow P(1):-8=−8$ which is true
Let P(k)be true,
$P(k):1^2−3^2+5^2−7^2+⋯+(4k-3)^2-(4k-1)^2=−8k^2$.......(1)
We have to prove P(k+1) as true.
$P(k+1):1^2−3^2+5^2−7^2+⋯+(4k-3)^2-(4k-1)^2+(4(k+1)-3)^2-(4(k+1)-1)^2=−8(k+1)^2$
$LHS=1^2−3^2+5^2−7^2+⋯+(4k-3)^2-(4k-1)^2+(4k+1)^2-(4k+3)^2$
$=-8k^2+(4k+1)^2-(4k+3)^2$
$=-8k^2+16k^2+8k+1-(16k^2+24k+9)$
$=-8k^2-16k-8=-8(k^2+2k+1)=-8(k+1)^2$=RHS
Therefore P(k+1) is true.So by Induction principle,P(n) is true for all $n$
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