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I need to prove that $$\sum\limits_{a=k}^\infty a2^{-a}\binom{a-1}{k-1}=2k$$ I have simplified this series to use the binomial theorem: $$\sum\limits_{b=0}^\infty (b+k)2^{-b-k}\binom{b+k-1}{k-1}$$ However, $(b+k)$ does not make that possible.

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$$\sum_{a=k}^{\infty}\color{blue}{a\binom{a-1}{k-1}}2^{-a}=\sum_{a=k}^{\infty}\color{blue}{k\binom{a}{k}}2^{-a}\underset{a=n+k}{=}2^{-k}k\sum_{n=0}^{\infty}\binom{n+k}{k}2^{-n}=2^{-k}k\left(1-\frac{1}{2}\right)^{-k-1}=2k.$$

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\on}[1]{\operatorname{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\sum_{a = k}^{\infty}a\,2^{-a} {a - 1 \choose k - 1}} = \sum_{a = k}^{\infty}a\,2^{-a}{a - 1 \choose a - k} \\[5mm] = &\ \sum_{a = k}^{\infty}a\,2^{-a} {-k \choose a - k}\pars{-1}^{a - k} \\[5mm] = &\ 2^{-k}\sum_{a = 0}^{\infty}\pars{a + k}\,\pars{-\,{1 \over 2}}^{a} {-k \choose a} \\[5mm] = &\ \left. 2^{-k}\pars{x\,\partiald{}{x} + k} \sum_{a = 0}^{\infty}{-k \choose a}x^{a} \,\right\vert_{\ x\ =\ -1/2} \\[5mm] = &\ \left. 2^{-k}\pars{x\,\partiald{}{x} + k} \pars{1 + x}^{-k} \,\right\vert_{\ x\ =\ -1/2} \\[5mm] = &\ 2^{-k}\,\bracks{\vphantom{\Large A}-kx\pars{1 + x}^{-k - 1}\ +\ k\pars{1 + x}^{-k}}_{\ x\ =\ -1/2} \\[5mm] = &\ \bbx{2k} \\ &\ \end{align}

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