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 $(2n+1)\cdot n$
 A: 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.
A: $$a_n=\sum_{k=1}^{2(n-1)}(-1)^kk^2+\sum_{k=2n-1}^{2n}(-1)^kk^2$$
$$=a_{n-1}+\sum_{k=2n-1}^{2n}(-1)^kk^2$$
$=2n^2-3n+1+(2n)^2-(2n-1)^2=\cdots$
W/O using induction,
$$a_n=\sum_{k=1}^{2n}(-1)^kk^2=\sum_{k=1}^n[(2k)^2-(2k-1)^2]=4\sum_{k=1}^nk-\sum_{k=1}^n1=4\cdot\dfrac{n(n+1)}2-n$$
A: Note: Here is a clear, clean way of going about proving your result going from $k$ to $k+1$:
For $n\geq 1$ let $S(n)$ denote the statement
$$
S(n) : \sum_{i=1}^{2n}(-1)^i\cdot i^2 = (2n+1)\cdot n.
$$
To prove this, we argue by induction on $n$. 
Base case ($n=1$): For $S(1)$, we have that
$$
\sum_{i=1}^{2}(-1)^i\cdot i^2=[(-1)^1\cdot 1^2]+[(-1)^2\cdot 2^2]=-1+4=3=(2 \cdot 1+1) \cdot 1. \quad\large\checkmark
$$
Inductive step ($S(k)\to S(k+1)$): Fix some $k\geq 1$ and suppose that
$$
S(k) : \sum_{i=1}^{2k}(-1)^i\cdot i^2 = (2k+1)\cdot k
$$
holds. To be shown is that
$$
S(k+1) : \sum_{i=1}^{2k+2}(-1)^i\cdot i^2 = (2k+3)(k+1)
$$
follows. Beginning with the left-hand side of $S(k+1)$,
\begin{align}
\sum_{i=1}^{2k+2}(-1)^i\cdot i^2 &= \sum_{i=1}^{2k}(-1)^i\cdot i^2 + [(-1)^{2k+2}\cdot(2k+2)^2] + [(-1)^{2k+1}\cdot(2k+1)^2]\quad\text{(by defn. of $\Sigma$)}\\[1em]
&= [(2k+1)\cdot k] + [4k^2+8k+4] - [4k^2+4k+1]\qquad\text{(by $S(k)$; simplify)}\\[1em]
&= [2k^2+k]+[4k+3]\qquad\text{(simplify)}\\[1em]
&= 2k^2+5k+3\qquad\text{(simplify)}\\[1em]
&= (2k+3)(k+1),\qquad\text{(factor)}
\end{align}
we see that the right-hand side of $S(k+1)$ follows, completing the inductive step. 
Thus, by mathematical induction, the statement $S(n)$ is true for all $n\geq 1$. $\blacksquare$
