<s>[INST] 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.
 [/INST] Well, if $n-1 =2k$ $$\displaystyle\sum_{i=1}^{n} \binom{i}{2} = \displaystyle\sum_{i=1}^{n}\frac{i\cdot (i-1)}{2}=\frac{1\cdot 0}{2}+\frac{2\cdot 1}{2}+\frac{3\cdot 2}{2}+\frac{4\cdot 3}{2} + \ldots +\frac{n\cdot (n-1)}{2} = \frac{2\cdot 1+3\cdot 2+4\cdot 3+ \ldots +n\cdot (n-1)}{2} = \frac{2^2\cdot2+4^2\cdot2+6^2\cdot2+\ldots +(n-1)^2\cdot2}{2}=2^2+4^2+6^2+\ldots+(n -1)^2=4(1+2^2+3^2+\ldots+k^2)=\frac{4k\cdot(k+1)\cdot(2k+1)}{6}=\frac{4\frac{n-1}{2}\cdot\frac{n+1}{2}\cdot n}{6} = \frac{(n+1)\cdot n\cdot(n-1)}{3!}=\binom{n+1}{3}$$
Similarly if $n-1 =2k +1$ you have $$\displaystyle\sum_{i=1}^{n} \binom{i}{2} = \displaystyle\sum_{i=1}^{n}\frac{i\cdot (i-1)}{2}=\frac{1\cdot 0}{2}+\frac{2\cdot 1}{2}+\frac{3\cdot 2}{2}+\frac{4\cdot 3}{2} + \ldots +\frac{n\cdot (n-1)}{2} = \frac{2\cdot 1+3\cdot 2+4\cdot 3+ \ldots +n\cdot (n-1)}{2} = \frac{2^2\cdot2+4^2\cdot2+6^2\cdot2+\ldots +(n-2)^2\cdot2+n\cdot(n-1)}{2}=2^2+4^2+6^2+\ldots+(n-2)^2+n\cdot (n-1)=4(1+2^2+3^2+4^2+\ldots+(n-2)^2)+\frac{n\cdot(n-1)}{2}=\frac{4k\cdot(k+1)\cdot(2k+1)}{6}+\frac{n\cdot(n-1)}{2}=\frac{4\frac{n-2}{2}\cdot\frac{n}{2}\cdot (n-1)}{6}+\frac{n\cdot(n-1)}{2}=\frac{(n-2)\cdot n\cdot(n-1)}{3!}+\frac{n\cdot(n-1)}{2}=\frac{n\cdot(n-1)\cdot(n-2)+3n\cdot(n-1)}{3!}=\frac{n\cdot(n-1)\cdot(n-2+3)}{3!}=\frac{n\cdot(n-1)\cdot(n+1)}{3!}=\binom{n+1}{3}$$
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