<s>[INST] 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!
 [/INST] Using your identity, we can write
$$\binom{k}{k} + \binom{k+1}{k} + \cdots + \binom{k+r}{k} = \binom{k+r+1}{k+1}$$
and take $k=2$ to give
$$\binom{2}{2} + \binom{3}{2} + \cdots + \binom{r+2}{2} = \binom{r+3}{3}$$ but we know that $\binom{k}{2} = \frac{k(k-1)}{2}$, so 
$$\begin{align}\binom{r+3}{3} &= \sum_{i=0}^r \binom{i+2}{2} \\
\frac{(r+3)(r+2)(r+1)}{6} &= \sum_{i=0}^r \frac{(i+1)(i+2)}{2} \\
&= \sum_{i=0}^r \left(\frac{1}{2}i^2 + \frac{3}{2}i+1\right) \\
&= \frac{1}{2}\sum_{i=1}^ri^2 + \frac{3}{2}\frac{r(r+1)}{2}+r+1
\end{align}$$
from which the desired identity easily follows.
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