Proof without using induction How to prove that 
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
without using induction.
If we don't know the right side of this expression, how to get right expression. I tried with partial sums and binomial formula but can't get it.
So the problem is:
$$1^2+2^2+...+n^2=?$$
Thanks for replies.

Assume we know $\sum\limits_{k=1}^nk=\frac{n(n+1)}{2}$. Compute the following cubes
$$\begin{align}
1^3&=1\\
(1+1)^3&=1^3+3\cdot 1^2+3\cdot 1+1^3\\
\vdots&=\vdots\\
n^3&=(n-1)^3+3(n-1)^2+3(n-1)+1^3\\
(n+1)^3&=n^3+3n^2+3n+1^3\end {align}$$
Add these equations together and cancel the cubes you have on both sides you get
$$\begin{align}(n+1)^3&=1+3\sum_{k=1}^nk^2+3\sum_{k=1}^nk+n\\
&=(n+1)\frac{3n+2}{2}+3\sum_{k=1}^nk^2\end{align}$$
This yields 
$$\sum_{k=1}^nk^2=\frac{n+1}{3}\left(n^2+2n+1-\frac{3n+2}{2}\right)=\frac{(n+1)(2n^2+n)}{6}$$
Factoring $n$ we get the result expected
