I have happened to have proved this sum while attempting to prove another summation. Let $$S_n=\sum_{k=0}^{2n} {2k \choose k } {2n \choose k}\left( \frac{-1}{2} \right)^k$$ ${2k \choose k} $ is the coefficient of $x^0$ in $(x+1/x)^{2k}$. Consequently, $S_n$ is the coefficient of $x^0$ in $$f(x)= \sum_{k=0}^{2n} {2n \choose k}~\left (x+\frac{1}{x}\right)^{2k}~\left ( \frac{-1}{2}\right)^{k}=\left(1-\left(\frac{x+1/x}{\sqrt{2}}\right)^2 \right)^{2n} = 4^{-n} ~ \left(x^2+\frac{1}{x^2}\right)^{2n}.$$ Finally, the coefficient of $x^0$ in $f(x)$ is $$4^{-n}~{2n \choose n}=S_n.$$ I hope that you will find it interesting and prove it in some other way. Do try!

  • $\begingroup$ Well, you made it in a very nice way :) $\endgroup$ – user657324 May 17 at 7:54
  • $\begingroup$ It's perfect. (+1) $\endgroup$ – user90369 May 17 at 10:19
  • $\begingroup$ Perhaps you should post your solution as an answer math.stackexchange.com/help/self-answer $\endgroup$ – leonbloy May 17 at 15:05
  • $\begingroup$ Unfortunately, not giving the solution with the problem can be dangerous. Yesterday, I posted the question: A family of lines normal to one and tangent to another curve. I got -2 for this. So I quickly posted its solution.today! $\endgroup$ – Dr Zafar Ahmed DSc May 19 at 2:44

We use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ of a series. Recalling the generating function of the central binomial coefficients we can write \begin{align*} [z^n]\frac{1}{\sqrt{1-4z}}=\binom{2n}{n}\tag{1} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{k=0}^{2n}}&\color{blue}{\binom{2n}{k}\binom{2k}{k}\left(-\frac{1}{2}\right)^k}\\ &=\sum_{k=0}^{2n}\binom{2n}{k}[z^k]\frac{1}{\sqrt{1+2z}}\tag{2}\\ &=[z^0]\frac{1}{\sqrt{1+2z}}\sum_{k=0}^{2n}\binom{2n}{k}z^{-k}\tag{3}\\ &=[z^0]\frac{1}{\sqrt{1+2z}}\left(1+\frac{1}{z}\right)^{2n}\tag{4}\\ &=[z^{-1}]\frac{(1+z)^{2n}}{z^{2n+1}\sqrt{1+2z}}\tag{5}\\ &=[t^{-1}]\frac{\left(1+\frac{t}{1-t}\right)^{2n}}{\left(\frac{t}{1-t}\right)^{2n+1}\sqrt{1+\frac{2t}{1-t}}}\cdot\frac{1}{(1-t)^2}\tag{6}\\ &=[t^{2n}]\frac{1}{\sqrt{1-t^2}}\tag{7}\\ &\,\,\color{blue}{=\frac{1}{4^n}\binom{2n}{n}}\tag{8} \end{align*}


  • In (2) we apply the coefficient of operator according to (1).

  • In (3) we use the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

  • In (4) we apply the binomial theorem.

  • In (5) we write the expression using formal residual by applying again the rule from comment (3).

  • In (6) we use the substitution $z=\frac{t}{1-t}, dz=\frac{1}{(1-t)^2}dt$.

  • In (7) we do some simplifications.

  • In (8) we select the coefficient of $t^{2n}$ by taking (1) evaluated at $z=\frac{1}{4}t^2$.


We will use $$ \frac1{1-x}\left(\frac{x}{1-x}\right)^k=\sum_{n=0}^\infty\binom{n}{k}x^n\tag1 $$ and $$ (1-4x)^{-1/2}=\sum_{k=0}^\infty\binom{2k}{k}x^k\tag2 $$ Extracting the even part of $(1)$ $$ \begin{align} \sum_{n=0}^\infty\binom{2n}{k}x^{2n} =\frac12\left[\frac1{1-x}\left(\frac{x}{1-x}\right)^k+\frac1{1+x}\left(-\frac{x}{1+x}\right)^k\right]\tag3 \end{align} $$ Compute the generating function of the sum we want $$ \begin{align} &\sum_{n=0}^\infty\sum_{k=0}^\infty\binom{2k}{k}\binom{2n}{k}\left(-\frac12\right)^kx^{2n}\tag4\\ &=\frac1{2-2x}\sum_{k=0}^\infty\binom{2k}{k}\left(-\frac{x}{2-2x}\right)^k +\frac1{2+2x}\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{x}{2+2x}\right)^k\tag5\\ &=\frac1{2-2x}\left(1+\frac{4x}{2-2x}\right)^{-1/2} +\frac1{2+2x}\left(1-\frac{4x}{2+2x}\right)^{-1/2}\tag6\\ &=\frac1{2-2x}\left(\frac{1-x}{1+x}\right)^{1/2} +\frac1{2+2x}\left(\frac{1+x}{1-x}\right)^{1/2}\tag7\\[6pt] &=\left(1-x^2\right)^{-1/2}\tag8\\[6pt] &=\sum_{n=0}^\infty\frac1{4^n}\binom{2n}{n}x^{2n}\tag9 \end{align} $$ Explanation:
$(4)$: compute the generating function
$(5)$: apply $(3)$
$(6)$: apply $(2)$
$(7)$: simplify
$(8)$: simplify
$(9)$: apply $(2)$

Equating coefficients of $x^{2n}$ gives $$ \sum_{k=0}^\infty\binom{2k}{k}\binom{2n}{k}\left(-\frac12\right)^k =\frac1{4^n}\binom{2n}{n}\tag{10} $$

  • $\begingroup$ Nice approach, nice presentation (+1) $\endgroup$ – Markus Scheuer May 23 at 13:33

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