<s>[INST] sum of binomial coefficients expansion to prove equation I want to prove that
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \binom{n+1}{3}
$$
I already expanded
$$
\binom{n+1}{3}
$$
to
$$
\binom{n+1}{3} = \frac{1}{6} * (n+1) *n*(n-1)
$$
and I know that the following equation must be right
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \frac{1}{6} * (n+1) *n*(n-1)
$$
but I do not get the expansion right, I tried starting with writing the sum explicit
$$
\binom{1}{2}+\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
since 1 over 2 is zero it can be shortened to
$$
\sum_{i=2}^{n}{\binom{i}{2}} = \binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
then I expanded the binomial coefficients to the corresponding factorial form
$$
\binom{n}{k} = \frac{n!}{k! * (n-k)!}
$$
but I do not get it right, could someone please help me?
 [/INST] $$\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}=\frac{1\times 2}{2}+\frac{2\times 3}{2}+\frac{3\times 4}{2}+...+\frac{(n-1)\times n}{2}=\frac{1}{2}(1\times 2+2\times 3+3\times 4+...+(n-1)\times n)=\frac {1}{2}\times \frac{(n-1)n(n+1)}{3}=\frac{(n-1)n(n+1)}{6}$$
</s>