2021/106) Evaluate closed form of $1^2+2^2+3^2+\cdots+n^2$

We have

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

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

Adding from 1 to n we get

$\sum_{k=1}^n((k+1)^3 - k^ 3) = 3\sum_{k=1}^n k^2 + 3\sum_{k=1}^nk + \sum_{k=1}^n1$

The LHS is a telespcopic sum = $(n+1)^3-1$

We know $\sum_{k=1}^nk = \frac{n(n+1)}{2}$

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

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

$=n^3 + 3n^2 + 3n + 1 - 3\frac{n^2+n}{2} - n$

$= \frac{1}{2}(2n^3 + 6n^2 + 6n - 3n^2 -3n -n )$

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

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

$= \frac{1}{2}n(n+2)(2n+1)$

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