Show that $ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$ I need a hand in showing that $$ \sum_{n=2}^m \binom{n}{2} = \binom{m+1}{3}$$ 
Thanks in advance for any help.

Using the Gosper's algorithm (Maxima command
AntiDifference(binomial(n,2),n)),
$$\binom{n}2 = \frac{(n+1)^3-3(n+1)^2+2(n+1)}6 - \frac{n^3-3n^2+2n}6$$
and the sum telescopes:
$$
\sum_{n=2}^m\binom{n}2 = \frac{(m+1)^3-3(m+1)^2+2(m+1)}6 =
\frac{(m+1)m(m-1)}{1\,2\,3}.
$$
