find the coefficients of $\frac{1}{x+k}$ in the summation

Question

i am able to prove the identity by putting the value of $A_n$ but cannot find it directly.I need help in finding the coefficents.

Answer 1

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\ds{\Large 13.\pars{\mrm{i}}:}$

• Multiply both members by $\pars{x + k}$: \begin{align} &{n! \over x\ldots \pars{x + k - 1}\pars{x + k + 1} \ldots\pars{x + n}} \\[2mm] = &\ A_{k} + \pars{x + k}\sum_{\substack{j = 0 \\ j \not= k}}^{n}{A_{j} \over x + j} \end{align}
• Takes the limit $\ds{x \to - k}$: \begin{align} &{n! \over \pars{-k}\ldots \pars{- 1}\pars{1} \ldots\pars{-k + n}} = A_{k} \\[5mm] & A_{k} = {n! \over k!\pars{-1}^{k}\pars{n - k}!} = \bbx{\pars{-1}^{k}{n \choose k}} \\ & \end{align}

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$\ds{\Large 13.\pars{\mrm{ii}}:}$ \begin{align} &\bbox[5px,#ffd]{\sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k}{1 \over \pars{k + 1}\pars{k + 2}}} \\[5mm] = &\ \sum_{k = 0}^{n}\pars{-1}^{k}{n \choose k} \int_{0}^{1}\pars{t^{k} - t^{k + 1}}\,\dd t \\[5mm] = &\ \int_{0}^{1}\bracks{% \sum_{k = 0}^{n}{n \choose k}\pars{-t}^{k} - t\sum_{k = 0}^{n}{n \choose k}\pars{-t}^{k}}\,\dd t \\[5mm] = &\ \int_{0}^{1}\pars{1 - t}\pars{1 - t}^{n}\,\dd t = \int_{0}^{1}t^{n + 1}\,\dd t = \bbx{1 \over n + 2} \\ & \end{align}