<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] For $m=2$, it amounts to proving $\binom{2}{2} = \binom{3}{3}$, which is true since both equal $1$.
Induction step: let's assume the formula is true for a given $m$,
$$\sum_{n=2}^m \binom{n}{2}=\binom{m+1}{3}$$
Then,
$$\sum_{n=2}^{m+1} \binom{n}{2} = \sum_{n=2}^{m} \binom{n}{2} + \binom{m+1}{2}$$
$$=\binom{m+1}{3}+\binom{m+1}{2}=\binom{m+2}{3}$$
And you are done, by induction.
By the way, the same reasoning would give you for any $k \ge 0$,
$$\sum_{n=k}^m \binom{n}{k} = \binom{m+1}{k+1}$$
For a better understanding of what it means, I suggest you draw Pascal's triangle and see the sum of binomial coefficients in a column, then the sum is at the bottom of this column, one step downward and one step to the right, as in the following:
$$ \begin{array}{cccccccc}
1 &   &   &   &   & & &\\
1 & 1 &   &   &   & & &\\
1 & 2 & \color{red}{1} &   &   & & &\\
1 & 3 & \color{red}{3} & 1 &   & & &\\
1 & 4 & \color{red}{6} & 4 & 1 & & &\\
1 & 5 & \color{red}{10} & 10 & 5 & 1& & &\\
1 & 6 & \color{red}{15} & 20 & 15 & 6& 1& &\\
1 & 7 & 21 & \color{blue}{35} & 35 & 21& 7& 1&\end{array}$$
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