combinatorics - 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?

Answer

$$\sum_{k=0}^{n-2}\frac{k^2+k(1-2n)+n^2-1}{2}\tag{Simplify what @dixv said}$$

now, $$\sum_{k=1}^nk=\frac{n(n+1)}{2}\text{ and }\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}{6}$$

Hope it helps?