<s>[INST] Proving $\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$ for all $n\geq 1$ by induction How prove the following equality:
$a_n$:=$\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$
$1$.presumption: $(-1)^1 \cdot 1^2+(-1)^2\cdot2^2=(2 \cdot 1+1) \cdot 1=3$ that seems legit
$2$.precondition:
$a_{n-1}$= $(2(n-1)+1)(n-1)$=$2n^2-3n+1$
for $k=n$
$a_n$= $\sum\limits_{k=1}^{2n} {(-1)^n \cdot n^2}$+$a_{n-1}$
$a_n$=${(-1)^n \cdot n^2}$+$2n^2-3n+1$
But that last equation seems somehow wrong to get $(2n+1)\cdot n$
 [/INST] $$a_n=\sum_{k=1}^{2(n-1)}(-1)^kk^2+\sum_{k=2n-1}^{2n}(-1)^kk^2$$
$$=a_{n-1}+\sum_{k=2n-1}^{2n}(-1)^kk^2$$
$=2n^2-3n+1+(2n)^2-(2n-1)^2=\cdots$
W/O using induction,
$$a_n=\sum_{k=1}^{2n}(-1)^kk^2=\sum_{k=1}^n[(2k)^2-(2k-1)^2]=4\sum_{k=1}^nk-\sum_{k=1}^n1=4\cdot\dfrac{n(n+1)}2-n$$
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