Question

While looking for a closed form of a expression I worked myself to a formula that resembles the Vandermonde convolution, but is summed over even binomial coefficients only.

$\sum_{k=0}^n\sum_{l=0}^n{{2k+2l}\choose{2l}}{{4n-2k-2l}\choose{2n-2l}}$

I'm at a loss as to what to do with it. I can re-write it in several ways, but the principal problem remains. Is there a known technique to attack such sums? Thanks.

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Answer 1

We have $$\sum_l\binom{a+l}lx^l=\frac1{(1-x)^{a+1}},$$ hence the generating function for the even terms of the sequence is $$\sum_l\binom{a+2l}{2l}x^{2l}=\frac12\left(\frac1{(1-x)^{a+1}}+\frac1{(1+x)^{a+1}}\right).$$ Consequently, \begin{multline*}\sum_l\binom{a+2l}{2l}\binom{b+2(n-l)}{2(n-l)}=\\ [x^{2n}]\frac14\left(\frac1{(1-x)^{a+1}}+\frac1{(1+x)^{a+1}}\right)\left(\frac1{(1-x)^{b+1}}+\frac1{(1+x)^{b+1}}\right),\end{multline*} where $[x^n]f$ denotes the $n$th coefficient of the power series for $f$. Plugging in the actual values for $a$ and $b$ and summing over $k$ gives \begin{align*} \sum_k\sum_l&\binom{2(k+l)}{2l}\binom{2(n-k)+2(n-l)}{2(n-l)}\\ &=[x^{2n}]\sum_{k=0}^n\frac14\left(\frac1{(1-x)^{2k+1}}+\frac1{(1+x)^{2k+1}}\right)\left(\frac1{(1-x)^{2(n-k)+1}}+\frac1{(1+x)^{2(n-k)+1}}\right)\\ &=[x^{2n}]\left[\frac{n+1}4\left(\frac1{(1-x)^{2n+2}}+\frac1{(1+x)^{2n+2}}\right)+\frac{1-x^2}{8x}\left(\frac1{(1-x)^{2n+2}}-\frac1{(1+x)^{2n+2}}\right)\right]\\ &=\frac{2n^2+4n+1}{2n+1}\binom{4n}{2n}=(2n^2+4n+1)C_{2n}. \end{align*}