Prove that $\sum_{i=2}^n \binom{n}{i} \binom{i}{2} = 2^{(n - 3)} \cdot n(n - 1)$ When attempting to solve an asymptotic complexity problem, i got this summation
$$\sum_{i=2}^n \binom{n}{i} \binom{i}{2}$$
Wolfram Alpha states that it equals to
$$2^{(n - 3)} \cdot n(n - 1)$$
But I can't find how, and Wolfram is unable to produce the step-by-step solution.

Consider
$$(1+x)^n=\sum_{i=0}^n\binom{n}ix^i.$$
Differentiate twice:
$$n(n-1)(1+x)^{n-2}=\sum_{i=2}^n\binom{n}ii(i-1)x^i.$$
Now set $x=1$.
