Proving $\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$ for all $n\geq 1$ by induction

Question

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 $(2n+1)\cdot n$

Answer 1

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.

Answer 2

$$a_n=\sum_{k=1}^{2(n-1)}(-1)^kk^2+\sum_{k=2n-1}^{2n}(-1)^kk^2$$

$$=a_{n-1}+\sum_{k=2n-1}^{2n}(-1)^kk^2$$

$=2n^2-3n+1+(2n)^2-(2n-1)^2=\cdots$

W/O using induction,

$$a_n=\sum_{k=1}^{2n}(-1)^kk^2=\sum_{k=1}^n[(2k)^2-(2k-1)^2]=4\sum_{k=1}^nk-\sum_{k=1}^n1=4\cdot\dfrac{n(n+1)}2-n$$

Answer 3

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 $n\geq 1$. $\blacksquare$