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.
Heres a nice combinatorial proof: Lets say you have $n+1$ kids, and want to form a committee of three. Order the kids $a_1, a_2, \cdots, a_{n+1}$. There are $\dbinom{n+1}{3}$ ways to form the committee. On the other hand, if $a_1$ is the first person on the committee, we need to choose two more, in $\dbinom{n}{2}$ ways. If $a_2$ is the first person on the committee, we can choose two more in $\dbinom{n-1}{2}$ ways. In general, if $a_n$ is the first person on the committee, we can choosse two more in $\dbinom{n-k+1}{2}$ ways. Therefore, we have $$ \dbinom{n+1}{3} = \sum_{i=1}^{n+1} \dbinom{n-k+1}{2} = \dbinom{n}{2} + \dbinom{n-1}{2} + \cdots + \dbinom22$$
If the identity is true, we can rewrite this specific case of the well-known addition formula 
$$\binom m3+\binom m2=\binom{m+1}3$$
as the trivial recurrence
$$\sum_{n=2}^{m-1} \binom{n}{2}+\binom m2=\sum_{n=2}^m \binom{n}{2}.$$