Proof $n \geq 3\hspace{0.5cm} \sum\limits_{k=1}^{n-2}{n-k\choose 2}={n\choose 3}$ How do you prove this expression trough mathematical induction ? It just ends up as a messy string of expressions when I try to solve it.
$n \geq 3\hspace{0.5cm} \sum\limits_{k=1}^{n-2}{n-k\choose 2}={n\choose 3}$

To prove:
$$\sum_{k=1}^{n-2} \binom{n-k}{2} = \binom{n}{3}$$
Base case ($n = 3$):
$$ \sum_{k=1}^{1} \binom{3-k}{2} = \binom{2}{2} = 1 = \binom{3}{3}
$$
Induction hypothesis:
For fixed $n$ we may assume: $\sum_{k=1}^{n-2} \binom{n-k}{2} = \binom{n}{3}$
Induction step ($n \to n+1$):  
\begin{align}
\sum_{k=1}^{n-1} \binom{n+1-k}{2} &= \binom{2}{2} + \sum_{k=1}^{n-2} \binom{n+1-k}{2} \\
&= 1 + \sum_{k=1}^{n-2} \left(\binom{n-k}{2} + n-k\right) \\
&= 1 + n(n-2) - \frac{(n-2)(n-1)}{2} +\sum_{k=1}^{n-2} \binom{n-k}{2} \\
&\overset{IH}{=} \frac{n(n-1)}{2} + \binom{n}{3} \\
&= \frac{n!}{2(n-2)!} + \binom{n}{3} \\
&= \binom{n}{2} + \binom{n}{3} \\
&= \binom{n+1}{3}
\end{align}
