Combinatorial sums with binomial coefficients

Question

Mathematica tells me that \begin{equation*} \sum_{k=a}^l \left(-\frac{1}{4}\right)^k {2k \choose k} {l \choose k} {k \choose a} = \frac{(-1)^a}{4^l} {2a \choose a} {2l - 2a \choose l-a}, \end{equation*} as well as \begin{equation*} \sum_{k=a}^l \left(-\frac{1}{4}\right)^k {2k-2a \choose k-a} {l \choose k} = \frac{(-1)^a}{4^l} \frac{{2l \choose l}{l \choose a}}{{2a \choose a}}. \end{equation*}

I would like to prove these results, but I don't know where to begin. Any help, hints or references are appreciated. Thanks!

Answer 1

Your sums are examples of hypergeometric series. If you write them in standard notation you can look them up in any book on special functions. The first one is $$\left(-\frac 14\right)^a\binom{2a}a\binom la\,{}_2F_1\left(\begin{matrix}a-l,a+1/2\\ a+1\end{matrix};1\right)$$ and the second one is $$\left(-\frac 14\right)^a\binom la\,{}_2F_1\left(\begin{matrix}a-l,1/2\\ a+1\end{matrix};1\right).$$ Since $l-a$ is a non-negative integer, these are special cases of the Chu-Vandermonde summation http://dlmf.nist.gov/15.4.E24 $${}_2F_1\left(\begin{matrix}-n,a\\ b\end{matrix};1\right)=\frac{(b-a)_n}{(b)_n},$$ where $(a)_n=a(a+1)\dotsm(a+n-1)$.

Answer 2

$$S=\sum_{k=a}^l \left(-\frac{1}{4}\right)^k {2k \choose k} {l \choose k} {k \choose a}$$ Denoting $m=l-a$ $$S=\frac{l!}{a!}\sum_{k=a}^l\left(-\frac{1}{4}\right)^k\frac{(2k)!}{(k!)^2}\frac1{(l-k)!(k-a)!}=\frac{l!}{a!}\sum_{k=0}^{l-a}\left(-\frac{1}{4}\right)^{a+k}\frac{(2(a+k)!}{((a+k)!)^2}\frac1{(l-a-k)!k!}$$ $$=\left(-\frac{1}{4}\right)^a\frac{l!}{a!m!}\sum_{k=0}^m\left(-\frac{1}{4}\right)^k\frac{(2(a+k)!)}{((a+k)!)^2}\frac{m!}{k!(m-k)!}$$ Using $\,\frac{(2(a+k)!)}{((a+k)!)^2}=\frac1{2\pi}\int_0^{2\pi}(1+e^{it})^{2(a+k)}e^{-it(a+k)}dt=\frac1{2\pi}\int_0^{2\pi}(2\cos\frac t2)^{2(a+k)}dt\,$ and performing summation $$S=\left(-\frac14\right)^a4^a\frac{l!}{a!m!}\frac1{2\pi}\int_0^{2\pi}\big(\cos\frac t2\big)^{2a}\big(1-\cos^2\frac t2\big)^mdt$$ $$=(-1)^a\frac{l!}{a!m!}\frac1\pi B\Big(a+\frac12;m+\frac12\Big)=(-1)^a\frac{l!}{a!(l-a)!}\frac1\pi\frac{\Gamma\Big(a+\frac12\Big)\Gamma\Big(l-a+\frac12\Big)}{\Gamma(l+1)}$$ $$=(-1)^a\frac1\pi\frac1{a!(l-a)!}\frac{\Gamma(a)\Gamma\Big(a+\frac12\Big)}{\Gamma(a)}\,\frac{\Gamma(l-a)\Gamma\Big(l-a+\frac12\Big)}{\Gamma(l-a)}$$ Using $\,\Gamma(z)\Gamma\big(z+\frac12\big)=2^{1-2z}\sqrt\pi\Gamma(2z)$ and $\,z\Gamma(z)=z!$ $$S=(-1)^a\frac{4\cdot2^{-2l}}{a!(l-a)!\Gamma(a)\Gamma(l-a)}\Gamma(2a)\Gamma\big(2(l-a)\big)=\frac{(-1)^a}{4^l} {2a \choose a} {2l - 2a \choose l-a}$$

Answer 3

First identity: $$ \boxed{\sum_{k} \left(-4\right)^{-k} {2k \choose k} {l \choose k} {k \choose a} = \frac{(-1)^a}{4^l} {2a \choose a} {2l - 2a \choose l-a}} $$ Note that $$ \binom{l}{k} \binom{k}{a} = \binom{l}{a} \binom{l-a}{l-k}, $$ so we can factor out $\binom{l}{a}$: $$ \binom{l}{a} \sum\limits_k (-4)^{-k} \binom{2k}{k} \binom{l-a}{l-k} $$

Now, we substitute $\binom{l-a}{l-k} = [x^l] x^k (1+x)^{l-a}$:

$$ \binom{l}{a} [x^l] (1+x)^{l-a} \sum\limits_k \binom{2k}{k} (-4)^{-k}x^k $$ Using the standard expansion $\frac{1}{\sqrt{1-4x}} = \sum\limits_k \binom{2k}{k} x^k$, we collapse the sum into $$ \binom{l}{a} [x^l] \frac{(1+x)^{l-a}}{\sqrt{1+x}} = \binom{l}{a} \binom{l-a-\frac{1}{2}}{l}, $$ which can be proven to be the same as RHS via Gamma function, or some "menial labor".

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Second identity:

$$ \boxed{\sum_{k} (-4)^{-k} {2k-2a \choose k-a} {l \choose k} = \frac{(-1)^a}{4^l} \frac{{2l \choose l}{l \choose a}}{{2a \choose a}}} $$ We can substitute $\binom{2k-2a}{k-a} =[x^{l-a}] x^{l-k} \frac{1}{\sqrt{1-4x}}$, which turns the sum into $$ [x^{l-a}]\frac{1}{\sqrt{1-4x}}\sum\limits_k \binom{l}{k} (-4)^{-k} x^{l-k}, $$ which, by binomial theorem, is equal to $$ [x^{l-a}] \frac{1}{\sqrt{1-4x}}\left(x-\frac{1}{4}\right)^l = (-4)^{-l}[x^{l-a}]\frac{(1-4x)^l}{\sqrt{1-4x}}=(-4)^{-a}\binom{l-\frac{1}{2}}{l-a}, $$ and some further manual verification again confirms that it is equal to the RHS.