-3

How can I find closed form formula for this summation:

  • $\sum_{k=0}^n k^2 \binom{n}{k} 3^{2k}$

Hordane
  • 1
  • Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. – Clement C. Feb 14 '16 at 22:59
  • 1
    This question has been asked many times before, but I can't find a duplicate right now. One way to solve it is to notice that the sum can be written $x\frac{d}{dx}\left[x\frac{d}{dx}f(x)\right]_{x=9}$ where $f(x) = \sum {n\choose k} x^k$. This sum can be evaluated by using the binomial theorem. – Winther Feb 14 '16 at 23:03
  • See this answer for a derivation – Winther Feb 14 '16 at 23:07

1 Answers1

2

$$(1+x)^n=\sum_{k=0}^n\binom nk x^k\implies n(1+x)^{n-1}=\sum_{k=1}\binom nkk\,x^{k-1}\implies$$

$$\implies n(n-1)(1+x)^{n-2}=\sum_{k=2}^n\binom nk k(k-1)x^{k-2}=\sum_{k=2}^n\binom nk k^2x^{k-2}-\sum_{k=2}^n\binom nk k\,x^{k-2}=$$

DonAntonio
  • 211,718
  • 17
  • 136
  • 287