Prove $\sum_{k=0}^{n-2}{n-k \choose 2} = {n+1 \choose 3}$ 
Prove 
  $$\sum_{k=0}^{n-2}{n-k \choose 2} = {n+1 \choose 3}$$

Is there a relation I can use that easily yields above equation?

Using index transformation $m = n - k$, reversal of the summation order and the definition of the binomial coefficient in terms of factorials we can write for $n\ge 2$:
\begin{align}
\sum_{k=0}^{n-2} \binom{n-k}{2}
&= \sum_{m=2}^{n} \binom{m}{2} \\
&= \sum_{m=2}^{n} \frac{m!}{2! (m-2)!} \\
&= \sum_{m=2}^{n} \frac{m(m-1)}{2} \\
&= \sum_{m=1}^n \frac{m(m-1)}{2} \\
&= \frac{1}{2} \sum_{m=1}^{n} m^2 - 
\frac{1}{2} \sum_{m=1}^n m \\
&= \frac{n(n+1)(2n+1)}{12} - \frac{n(n+1)}{4} \\
&= \frac{1}{6} n^3 + \frac{1}{4} n^2 + \frac{1}{12} n -
\left( \frac{1}{4} n^2 + \frac{1}{4} n \right) \\
&= \frac{1}{6} n^3 - \frac{1}{6} n
\end{align}
where we used Faulhaber's formula for the square pyramidal and triangular numbers. 
On the other side of the equation we have:
\begin{align}
\binom{n+1}{3} 
&= \frac{(n+1)!}{3!(n+1-3)!} \\
&= \frac{(n+1)n(n-1)}{6} \\
&= \frac{n^3 - n}{6}
\end{align}
