<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] As $\displaystyle\binom n2=\frac{n(n-1)}2=\frac12\cdot n^2-\frac12\cdot n$
$$\sum_{2\le n\le m}\binom n2=\frac12 \sum_{2\le n\le m}n^2-\frac12\sum_{2\le n\le m} n$$
$$=\frac12\left( \sum_{1\le n\le m}n^2-1\right)-\frac12\left(\sum_{1\le n\le m} n-1\right)$$
$$=\frac12 \frac{m(m+1)(2m+1)}6-\frac12\frac{m(m+1)}2$$
$$=\frac{m(m+1)(2m+1)-3m(m+1)}{12}$$
$$=\frac{(m+1)m}{12}(2m+1-3)=\frac{(m+1)m(m-1)}6=?$$
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