Proving $\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$ without induction I was looking at: $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$
It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation?

$\displaystyle\sum_{k=0}^{n} k^{3} + (n+1)^{3} = 0^{3} + \sum_{k=1}^{n+1}k^{3} = \sum_{k=0}^{n} (k+1)^{3} = \sum_{k=0}^{n} (k^{3} +3k^{2}+3k+1)$
$\displaystyle = \sum_{k=0}^{n} k^{3} + \sum_{k=0}^{n} k^{2} + 3 \frac{n(n+1)}{2} + (n+1)$
$ \displaystyle \implies 3 \sum_{k=0}^{n} k^{2} = 3 \sum_{k=1}^{n} k^{2} = (n+1)^{3} -3 \frac{n(n+1)}{2} -(n+1) = \frac{n(n+1)(2n+1)}{2}$
Similarly, the  sum $\displaystyle \sum_{k=1}^{n} k^{3} $ can be evaluated by starting with the equation $\displaystyle\sum_{k=0}^{n} k^{4} + (n+1)^{4}= 0^{4} + \sum_{k=1}^{n+1} k^{4}$
