<s>[INST] Combinatorial proof of $\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$ and then use result to find a formula for $1^2 + 2^2 + \ldots + n^2$ a) Give a combinatorial proof that for every $n \geq r \geq 1$ that:
$$\sum_{i = 0}^{n} \binom{i}{r - 1} = \binom{n + 1}{r}$$
And use (a) to concoct a formula for $1^2 + 2^2 + ... + n^2$
 [/INST] Maybe not combinatorial, but maybe it helps somebody else for a proper idea.
\begin{align}
\binom{n+1}{r+1} &= \frac{n+1}{r+1}\binom{n}{r} \\
&=\binom{n}{r} + \frac{n-r}{r+1} \binom{n}{r} =\binom{n}{r} + \frac{n}{r+1} \binom{n-1}{r} =\binom{n}{r} + \binom{n}{r+1} \\
&=\binom{n}{r} + \binom{n-1}{r} + \binom{n-1}{r+1} \\
&= \dots \\
&=\sum_{i=0}^k \binom{n-i}{r} + \binom{n-k}{r+1} \, .
\end{align}
The last term vanishes for $k=n-r$. In that case use this formula for $r=2$ which immediately yields $1^2 + 2^2 + \dots + n^2$ in terms of simpler sums.
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