Recently, I came across the following exercise on the course of discrete math

Find a closed form for $\sum_{k=0}^nk\binom{k}{3}\binom{2n}{k}$

So I tried some of the usual techniques: Let $f(x)=\sum_{k=0}^n\binom{2n}{k}x^k$.
If we denote $F(x)=\frac16x^3\left(xf'(x)\right)'''$ then $F(x)=\sum_{k=0}^nk\binom{k}{3}\binom{2n}{k}x^k$, so the desired sum is $F(1)$.
Problem here is that I can't find a closed expression for $f(x)$. Clearly, $(1+x)^{2n}=\sum_{k=0}^{2n}\binom{2n}{k}x^k$, but how can I take out only the first 'half' of the terms?

Second thing I tried was to take $g(x)=(1+x)^{2n}$ and then denote $G(x)=\frac16x^3\left(xg'(x)\right)'''$.
Then $G(x)=\sum_{k=0}^{2n}k\binom{k}{3}\binom{2n}{k}x^k$. So, if we look at $$h(x)=G(x)\frac1{1-x}=\sum_{j=0}^\infty\left(\sum_{k=0}^jk\binom{k}{3}\binom{2n}{k}\right)x^j$$ So the desired sum will be the coefficient of $x^n$. But here it doesn't work either, as
$$h(x)=\binom{2n}{3}(2nx+3)\frac{(1+x)^{2n-4}}{1-x}$$ But I can't find a closed form for the coefficient of $x^n$.

Any help will be much appreciated.


$$\sum_{k=0}^{n} k{k\choose 3}{2n\choose k}\\ =\sum_{k=0}^{n}\frac{k( k!)(2n)!}{(k-3)!3!k!(2n-k)!}\\ ={2n\choose 3}\sum_{k=0}^{n}\frac{k(2n-3)!}{(k-3)!(2n-k)!}\\ ={2n\choose 3}\sum_{k=0}^{n}k{2n-3\choose k-3}\\ ={2n\choose 3}\sum_{h=0}^{n-3}(h+3){2n-3\choose h}\\ ={2n\choose 3}\sum_{h=0}^{n-3}\left[(2n-3){2n-4\choose h-1}+3{2n-3\choose h}\right]\\ ={2n\choose 3}\left[(2n-3)\left(2^{2n-5}-{2n-4\choose n-2}/2-{2n-4\choose n-3}\right)\\ +3(2^{2n-4}-{2n-3\choose n-2})\right]$$

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  • $\begingroup$ How did you get the last equality? $\endgroup$ – Dennis Gulko Jul 5 '14 at 8:21
  • 2
    $\begingroup$ He is exploting the facts that $$\sum_{h=0}^{n}\binom{2n+1}{h}=2^{2n},\quad \sum_{h=0}^{n}\binom{2n+2}{h}=\frac{2^{2n+2}-\binom{2n+2}{n+1}}{2}.$$ $\endgroup$ – Jack D'Aurizio Jul 5 '14 at 9:58

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