Prove that $1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$ without using induction. I have to deduce the following formula $$1^2 + 2^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6},$$ while using the given formula $$\binom{k}{0}+\binom{k+1}{1}+\cdots+\binom{k+r}{r}=\binom{k+r+1}{r}$$
I tried to find values for $k$, such that $\binom{k}{0}=1^2$ etc. but that didn't work. Does anybody have a push in the right direction? Thanks!

When I was (much) younger I wanted to find the area of a segment of parabola so I needed the sum of the squares and I found the formula by myself in this way. 
Inspired by the famous $1+2+\ldots+n=\dfrac{n(n+1)}{2}$
I supposed that the sum of the square could be a third degree polynomial in $n$
$P(n)=an^3+bn^2+cn$ 
Then I plugged the first $3$ values for $n$ getting
$
\left\{
\begin{array}{l}
 a+b+c=1 \\
 8a+4b+2c=5 \\
 27 a + 9 b + 3 c=14\\
\end{array}
\right.
$
which gives $a=\frac13;\;b=\frac12;\;c=\frac16$
and therefore
$$P(n)=\frac13 n^3+\frac12 n^2+\frac16  n=\frac16 (2 n^3+3 n^2+n)=\frac16 n(n+1)(2n+1)$$
It is not elegant, but in 1977 there was no wikipedia :)
Hope it can be useful
