<s>[INST] 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.
 [/INST] A slight generalization:
$$
\begin{align}
\sum_{k=0}^n\binom{k}{a}\binom{n-k}{b}
&=\sum_{k=0}^n\binom{k}{k-a}\binom{n-k}{n-k-b}\tag{1}\\
&=\sum_{k=0}^n(-1)^{n-a-b}\binom{-a-1}{k-a}\binom{-b-1}{n-k-b}\tag{2}\\
&=(-1)^{n-a-b}\binom{-a-b-2}{n-a-b}\tag{3}\\
&=\binom{n+1}{n-a-b}\tag{4}\\
&=\binom{n+1}{a+b+1}\tag{5}
\end{align}
$$
$(1)$: $\binom{n}{k}=\binom{n}{n-k}$
$(2)$: $\binom{n}{k}=(-1)^k\binom{k-1-n}{k}$
$(3)$: Vandermonde's Identity
$(4)$: $\binom{n}{k}=(-1)^k\binom{k-1-n}{k}$
$(5)$: $\binom{n}{k}=\binom{n}{n-k}$  

Use the identity above to show that
$$
\sum_{n=0}^m\binom{n}{2}\binom{m-n}{0}=\binom{m+1}{2+0+1}=\binom{m+1}{3}
$$
</s>