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Hey guys can't find to figure this one out

$\sum_k^n k^{2}\left(\begin{array}{c}n\\ k\end{array}\right) = n(n+1)2^{n-2} ,k\geq0$

Maybe one of you can help.

Ajax Edm
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    My guess is that this is a straightforward induction proof. – B. Goddard Sep 29 '16 at 14:22
  • Hmm how would i atempt it on this one ? – Ajax Edm Sep 29 '16 at 14:24
  • Look at the first two derivatives of $(1+x)^n$ and try to write $\sum_{k=0}^n k^2\binom nk x^k$ in as linear combination of them. –  Sep 29 '16 at 14:24
  • The problem is we have not introduced the derivative yet in class so we are not allowed to use it – Ajax Edm Sep 29 '16 at 14:27
  • If you're in a discrete mathematics course it's unlikely that they will "introduce" a derivative to you there. Unless the instructor explicitly said inductive proof or combinatorial proof, I'd expect it to be fair game. You might want to look at taking derivatives of $(1+x)^{n+1}$ (twice), then expanding using the binomial theorem and evaluating at $x=1$. – TravisJ Sep 29 '16 at 14:36
  • Duplicate: http://math.stackexchange.com/questions/1431112/what-is-the-sum-sum-k-0nk2-binomnk

    http://math.stackexchange.com/questions/355262/closed-form-expression-for-sum-k-0n-binomnkkp-for-integers-n-p

    http://math.stackexchange.com/questions/545879/how-to-prove-that-sum-k-0n-binom-nk-k2-2n-2n2n

     – Darío A. Gutiérrez Sep 29 '16 at 14:43

1 Answers1

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$$\sum k^{2}\binom{n}{k}=\sum k\left(k-1\right)\binom{n}{k}+\sum k\binom{n}{k}=n\left(n-1\right)\sum\binom{n-2}{k-2}+n\sum\binom{n-1}{k-1}=$$$$n\left(n-1\right)2^{n-2}+n2^{n-1}=n\left(n+1\right)2^{n-2}$$

Using the convention that $\binom{n}{k}=0$ if $k\notin\{0,\dots,n\}$ just let $k$ range over $\mathbb Z$.

drhab
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  • Why is this true $\sum_k^n k\left(\begin{array}{c}n\ k\end{array}\right) = n\sum_k^n\left(\begin{array}{c}n-1\ k-1\end{array}\right)$ Is there any proof cause we didn't discuss it in class – Ajax Edm Sep 29 '16 at 14:45
  • $k\binom{n}{k}=k\frac{n!}{k!(n-k)!}=\frac{n!}{(k-1)!(n-k)!}=n\frac{(n-1)!}{(k-1)!(n-k)!}=n\binom{n-1}{k-1}$ – drhab Sep 29 '16 at 14:49
  • Ahh thanks man i get it ;) That makes sense now :) Thank's for the answer – Ajax Edm Sep 29 '16 at 14:50
  • You are welcome. – drhab Sep 29 '16 at 14:52