<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] The equality is true for $n=1$, because
$$
(-1)^1\cdot 1^2+(-1)^2\cdot 2^2=-1+4=3
$$
and
$$
(2\cdot 1+1)\cdot 1=3.
$$
Now suppose the assert is true for $a_{n-1}$. Then
\begin{align}
a_{n}&=\sum_{k=1}^{2n}(-1)^k\cdot k^2\\
&=\biggl(\sum_{k=1}^{2n-2}(-1)^k\cdot k^2\biggr)+
(-1)^{2n-1}\cdot(2n-1)^2+(-1)^{2n}(2n)^2\\
&=a_{n-1}+(-1)^{2n-1}\cdot(2n-1)^2+(-1)^{2n}(2n)^2\\
&=(2(n-1)+1)(n-1)-(2n-1)^2+(2n)^2
\end{align}
(the last equality by the induction hypothesis).
You have to verify that the last expression equals
$$
(2n+1)\cdot n
$$
which is easy algebra.
You seem to have forgotten a summand.
</s>