How to show $\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$? 
Show that $\,\displaystyle\sum_{i=1}^{n} \binom{i}{2}=\binom{n+1}{3}$.

I'm thinking right now (though not getting anywhere with it) that I want to expand out the summation portion to $i!/2!(i-2)!$ and simplify from there? Not sure if that will help, not to mention if I put $1$ in for $i$ I get $1/-2$ which I don't think is right. 
Anyone care to shed some light on the subject? 
Thanks.

$$\sum_{i=2}^n\binom{i}{2}=\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
Since $\binom22=\binom33$
$$\sum_{i=2}^n\binom{i}{2}=\binom{3}{3}+\binom{3}{2}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
Pascal's Triangle Identity states that
$$\binom{n-1}k+\binom{n-1}{k-1}=\binom{n}{k}$$
So,
$$\sum_{i=2}^n\binom{i}{2}=\color{red}{\binom{3}{3}+\binom{3}{2}}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
$$\sum_{i=2}^n\binom{i}{2}=\color{red}{\binom{4}{3}}+\binom{4}{2}+\binom{5}{2}+\cdots+\binom{n}{2}$$
$$\sum_{i=2}^n\binom{i}{2}=\color{green}{\binom{4}{3}+\binom{4}{2}}+\binom{5}{2}+\cdots+\binom{n}{2}$$
$$\sum_{i=2}^n\binom{i}{2}=\color{green}{\binom{5}{3}}+\binom{5}{2}+\cdots+\binom{n}{2}$$
So on and so forth, until we have
$$\sum_{i=2}^n\binom{i}{2}=\binom{n}{3}+\binom{n}{2}=\binom{n+1}{3}$$
