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?
 A: The right hand side ${n+1 \choose 3}$ is the number of ways to choose 3 elements from $\{1,2,\ldots,n+1\}$. Let $A$ denote the set of all 3-subsets of $\{1,2,\ldots,n+1\}$.  We want to show that $|A|$ is equal to the left hand side.  
Let $A_i$ denote the set of all 3-subsets of $\{1,2,\ldots,n+1\}$ which have $i$ as their largest element.  For example, if $i=n+1$, then $A_{n+1}$ is the set of all 3-subsets which contain $n+1$, and the number of ways to choose the remaining 2 elements is ${n \choose 2}$. Hence,  $|A_{n+1}| = {n \choose 2}$.  More generally, $|A_i| = {i-1 \choose 2}$, for $i=3,\ldots,n+1$, because the remaining 2 elements must be chosen from $\{1,\ldots,i-1\}$.   Note that $i$ has to be at least 3 because the largest of 3 elements will be at least 3.
Using the fact that the $A_i$'s $(i=3,4,\ldots,n+1)$ are disjoint and their union is all of $A$, we obtain the desired formula.
A: $$\sum_{k=0}^{n-2}\frac{k^2+k(1-2n)+n^2-1}{2}\tag{Simplify what @dixv said}$$
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?
A: For $$S=\sum_{k=0}^{n-2}(n-k)(n-k-1)$$  set $n-k=r$ to get $$S=\sum_{r=1}^nr(r-1)=\sum_{r=1}^nr^2-\sum_{r=1}^nr$$
Now use this  OR How to get to the formula for the sum of squares of first n numbers?
A: 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}
A: You can prove it using generating functions. We wish to prove that
$$
\sum_{m=0}^{n} \binom{m}{2}=\binom{n+1}{3}\tag{1}
$$
Use the identity 
$$
\frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{k+n-1}{k-1}x^n.\tag{2}
$$
(which can be obtained by repeatedly differentiating the geometric series) where $k\geq1$ to get that the generating function whose $m$th coefficent is $\binom{m+2}{2}$ is $(1-x)^{-3}$ and the generating function whose $m$th coefficent is $\binom{m}{2}$ is
$x^2(1-x)^{-3}$. Thus
$$
\sum_{m=0}^{n} \binom{m}{2}=
[x^n]\left(\frac{1}{1-x}\frac{x^2}{(1-x)^{3}}\right)
= [x^n] \left(\frac{x^2}{(1-x)^{4}}\right)\tag{3}
=\binom{n+1}{3}
$$
by (2) as desired where $[x^n]$ extracts the coefficient of $x^n$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{k = 0}^{n - 2}{n - k \choose 2} & =
\sum_{k = 0}^{n - 2}{k + 2\choose 2} = \sum_{k = 2}^{n}{k \choose 2}=
\sum_{k = 2}^{n}{k^{\underline{2}} \over 2} =
{1 \over 2}\left.{k^{\underline{3}} \over 3}\,\right\vert_{\ 2}^{\ n + 1} =
{\pars{n + 1}^{\underline{3}} - 2^{\underline{3}}\over 6}
\\[5mm] & = {\pars{n + 1}n\pars{n - 1} \over 6}
\end{align}
A: With so many proofs there is no need for one more. But I like using finite differences. 
Let $f(n)$ represent the sum. 
$\Delta f(n) = \binom {n+1}{2} = \Delta \binom {n+1}{3}$
$ \implies f(n) = \binom {n+1}{3} + c$
for $n=2$, we get $c + 1 = 1 \implies c=0$
