Show that $\sum_{k=1}^n k^2 {n\choose k} = n(n+1)2^{n-2}$.

Question

I'm trying to figure out how to prove that $$\sum_{k=1}^n k^2 {n\choose k} = n(n+1)2^{n-2}$$ I suspect that it has to do with using $$2^{n-2} = \sum_{k=1}^{n-1} {n-2 \choose k-1}$$ But I'm really not sure. Any help would be appreciated, thanks!

You should maybe consider to search here on MSE before asking a question since your last question - which is by the way quite similiar to this one - was a duplicate as well. — mrtaurho, Nov 04 '18 at 19:06
Are you aware about derivatives or you need to derive it without that? — user, Nov 04 '18 at 19:11
Are you going to ask about $$\sum_{k=1}^n k^3\binom nk$$ tomorrow? — Angina Seng, Nov 04 '18 at 19:22

score 2 · Answer 1 · answered Nov 04 '18 at 19:08

2

HINT

We have that

$$(1+x)^n=\sum_{k=0}^n x^k\binom{n}{k}$$

then derive twice and set a convenient value for $x$.

answered Nov 04 '18 at 19:08

user

154,566

Show that $\sum_{k=1}^n k^2 {n\choose k} = n(n+1)2^{n-2}$.

1 Answers1