The sum of the squares of the first n natural numbers

What is the sum of the squares of the first n natural numbers?

The answer is:

$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}$

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n:

$1^2+2^2+...+n^2$ :

Proof:

$(k+1)^3=k^3+3k^2+3k+1$

We write the above identity for k from 1 to n:

$2^3=1^3+3\cdot 1^2+3\cdot 1+1$
$3^3=2^3+3\cdot 2^2+3\cdot 2+1$
$4^3=3^3+3\cdot 3^2+3\cdot 3+1$
$5^3=4^3+3\cdot 4^2+3\cdot 4+1$
...
$(n+1)^3=n^3+3\cdot n^2+3\cdot n+1$

We add this n identities and we get:

$(n+1)^3=1^3+3\sum_{k=1}^{n} k^2+3\sum_{k=1}^{n} k+\sum_{k=1}^{n}1$

$n^3+3n^2+3n+1=1+3\sum_{k=1}^{n} k^2+3\frac{n(n+1)}{2}+n$

$n^3+3n^2+3n=3\sum_{k=1}^{n} k^2+3\frac{n(n+1)}{2}+n$

$n^3+3n^2+2n - 3\frac{n(n+1)}{2}=3\sum_{k=1}^{n} k^2$

$n(n^2+3n+2) - 3\frac{n(n+1)}{2}=3\sum_{k=1}^{n} k^2$

$n(n+1)(n+2) - 3\frac{n(n+1)}{2}=3\sum_{k=1}^{n} k^2$

$n(n+1)(n+2-\frac{3}{2})=3\sum_{k=1}^{n} k^2$

$n(n+1)(\frac{2n+1}{2})=3\sum_{k=1}^{n} k^2$

$\frac{n(n+1)(2n+1)}{6}=\sum_{k=1}^{n} k^2$