finding complex binomial sum in closed form

Question

Find sum of $\displaystyle \sum^{n}_{k=0}(-1)^k2^{2k}\binom{n}{k}\binom{2(n-k)}{n-k}$

Attempt: Coefficients of $x^k$ in $(1+x)^k$ and coefficients of $x^{n-k}$ in $(1+x)^{2(n-k)}$

could some help me how to solve it, thanks

Answer 1

We use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g. \begin{align*} [z^k](1+z)^n=\binom{n}{k} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{k=0}^n}&\color{blue}{(-1)^k2^{2k}\binom{n}{k}\binom{2n-2k}{n-k}}\\ &=\sum_{k=0}^\infty(-1)^k2^{2k}[z^k](1+z)^n[u^{n-k}](1+u)^{2n-2k}\tag{1}\\ &=[u^n](1+u)^{2n}\sum_{k=0}^\infty(-1)^k2^{2k}\frac{u^k}{(1+u)^{2k}}[z^k](1+z)^n\tag{2}\\ &=[u^n](1+u)^{2n}\sum_{k=0}^\infty\left(-\frac{4u}{(1+u)^2}\right)^k[z^k](1+z)^n\tag{3}\\ &=[u^n](1+u)^{2n}\left(1-\frac{4u}{(1+u)^2}\right)^n\tag{4}\\ &=[u^n](1-u)^{2n}\tag{5}\\ &\color{blue}{=(-1)^n\binom{2n}{n}} \end{align*}

Comment:

• In (1) we apply the coefficient of operator twice and we extend the upper limit of the sum to $\infty$ without changing anything, since we are adding zeros only.

• In (2) we use the linearity of the coefficient of operator and apply the rule \begin{align*} [z^{p-q}]A(z)=[z^p]z^qA(z) \end{align*}

• In (3) we apply the substitution rule of the coefficient of operator with $z:=-\frac{4u}{(1+u)^2}$ \begin{align*} A(u)=\sum_{n=0}^\infty a_nu^n=\sum_{n=0}^\infty u^n[z^n]A(z) \end{align*}

• In (4) we do some simplifications.

• In (5) we select the coefficient of $u^{n}$.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Hereafter, I'll use the identity $\ds{{2m \choose m} = {-1/2 \choose m}\pars{-4}^{m}}$.

\begin{align} &\sum_{k=0}^{n}\pars{-1}^{k}\,2^{2k}{n \choose k}{2\bracks{n - k} \choose n - k} = \sum_{k=0}^{n}\pars{-1}^{k}\,2^{2k}{n \choose k}{-1/2 \choose n - k} \pars{-4}^{n - k} \\[5mm] = &\ \pars{-4}^{n}\sum_{k=0}^{n}{n \choose k}{-1/2 \choose n - k} = \pars{-1}^{n}\,2^{2n}\sum_{k=0}^{n}{n \choose k} \bracks{z^{n - k}}\pars{1 + z}^{-1/2} \\[5mm] = &\ \pars{-1}^{n}\,2^{2n} \bracks{z^{n}}\bracks{\pars{1 + z}^{-1/2}\sum_{k=0}^{n}{n \choose k}z^{k}} = \pars{-1}^{n}\,2^{2n}\bracks{z^{n}}\pars{1 + z}^{n - 1/2} \\[5mm] = &\ \pars{-1}^{n}\,2^{2n}{n - 1/2 \choose n} = \pars{-1}^{n}\,2^{2n}{-1/2 \choose n}\pars{-1}^{n} = 2^{2n}{2n \choose n}\pars{-4}^{-n} = \bbx{\ds{\pars{-1}^{n}{2n \choose n}}} \end{align}

Answer 3

I would like to contribute to the collection, the challenge being that we use a method that is different from the two that were already presented. We seek to evaluate

$$\sum_{k=0}^n (-1)^k 2^{2k} {n\choose k} {2(n-k)\choose n-k}.$$

We use the integral

$${2n-2k\choose n-k} = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n-k+1}} \frac{1}{(1-z)^{n-k+1}} \; dz.$$

We obtain for the sum

$$\frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}} \sum_{k=0}^n {n\choose k} (-1)^k 2^{2k} z^k (1-z)^k \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}} (1-4z(1-z))^n \; dz \\ = \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} \frac{1}{(1-z)^{n+1}} (1-2z)^{2n} \; dz.$$

This is

$$\sum_{q=0}^n (-1)^q 2^q {2n\choose q} {2n-q\choose n}.$$

We have

$${2n\choose q} {2n-q\choose n} = \frac{(2n)!}{q! n! (n-q)!} = {2n\choose n} {n\choose q}$$

so this becomes

$${2n\choose n} \sum_{q=0}^n (-1)^q 2^q {n\choose q} = {2n\choose n} (1-2)^n $$

which is

$$\bbox[5px,border:2px solid #00A000]{ (-1)^n {2n\choose n}.}$$