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!

(Not using the combinatorial identity but without induction) 
You can start saying that $$(k+1)^{3}-k^3=3k^2+3k+1$$
then you take the sum in both sides to get:
$$\sum_{k=1}^{n}\bigg((k+1)^{3}-k^3\bigg)=\sum_{k=1}^{n}\big(3k^2+3k+1\big)$$
The LHS is a telescopic sum and RHS will give you the sum you are looking for:
$$ (n+1)^3-1^3=\sum_{k=1}^{n}\bigg((k+1)^{3}-k^3\bigg)=3\sum_{k=1}^{n}k^2+3\sum_{k=1}^{n}k+\sum_{k=1}^{n}1 
=3\sum_{k=1}^{n}k^2+3\frac{n(n+1)}{2}+n$$
and this implies:
$$ \frac{(n+1)^3-1-n-3\frac{n(n+1)}{2}}{3}=\sum_{k=1}^{n}k^2 $$
You can compute $\sum_{k=1}^{n}k$ without induction by saying that $$(k+1)^{2}-k^{2}=2k+1$$ and repeat the last argument. 
