Formula $\sum_{k=0} ^n \binom{n}{k}\cdot k^2 $

Question

I'm looking to generalize this to a formula with only n. I have figured out that $\sum_{k=0} ^n \binom{n}{k}\cdot k = n \cdot 2^{n-1}$ but I'm not quite sure how to leverage this to find this formula

Answer 1

We have the convenient formula $\binom nk = \frac nk \cdot \binom{n-1}{k-1} = \frac nk \cdot \frac{n-1}{k-1} \cdot \binom{n-2}{k-2}$.

This makes it easier to compute $$ \sum_{k=0}^n k(k-1) \binom nk $$ because we can rewrite this sum (skipping the zero terms for $k=0$ and $k=1$) as $$ \sum_{k=2}^n k(k-1) \frac{n(n-1)}{k(k-1)} \binom{n-2}{k-2} = n(n-1) \sum_{k=2}^n \binom{n-2}{k-2} $$ (which you should already know how to simplify).

But we can express the sum of $k^2 \binom nk$ as the sum of $k(k-1) \binom nk$ plus the sum of $k \binom nk$.