Closed Form For Summation

Question

$$\sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k}$$

What is the closed form (given $n$ is even)?

I've got to $n(2^2\binom{n-1}{1}+2^4\binom{n-1}{3}+\cdots+2^n\binom{n-1}{n-1})$, but I can't simplify further. Please help.

Answer 1

Hint: For every nonnegative integer $n$, we have $n\,x(1+x)^n=\sum\limits_{j=1}^n\,j\,\binom{n}{j}\,x^j$. Try plugging in $x:=\pm2$ and show that the answer is $n\,\big(3^{n-1}-(-1)^{n-1}\big)$.

Answer 2

$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} \color{#f00}{\sum_{k = 0}^{n/2}2^{2k}\pars{2k}{n \choose 2k}} & = \sum_{k = 0}^{n}2^{k}\,k{n \choose k}\,{1 + \pars{-1}^{k} \over 2} = \half\sum_{k = 0}^{n}{n \choose k}k\,2^{k} + \half\sum_{k = 0}^{n}{n \choose k}k\,\pars{-2}^{k}\label{1}\tag{1} \end{align}

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However, $$ \sum_{k = 0}^{n}{n \choose k}k\,x^{k} = x\,\partiald{}{x}\sum_{k = 0}^{n}{n \choose k}x^{k} = x\,\partiald{\pars{1 + x}^{n}}{x} = nx\pars{1 + x}^{n - 1} $$

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With this expression, \eqref{1} becomes \begin{align} \color{#f00}{\sum_{k = 0}^{n/2}2^{2k}\pars{2k}{n \choose 2k}} & = \half\,n\times\pars{2}\pars{1 + 2}^{\, n - 1} + \half\,n\times\pars{-2}\bracks{1 + \pars{-2}}^{\, n - 1} \\[5mm] & = \color{#f00}{n\bracks{3^{n - 1} + \pars{-1}^{n}}} \end{align}

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However, because n is even the expression can simplify to n[3^(n−1)+1]