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The question is : " proof for any N that -" $$\sum _{k=1}^n\:\left(-1\right)^{k+1}\cdot \left(2k-1\right)=\left(-1\right)^{n+1}\cdot n$$ I`ve reached to $$\left(-1\right)^{n+1}\cdot n+\left(-1\right)^{n+2}\cdot \left(2n+1\right)=\left(-1\right)^{n+2}\cdot \left(n+1\right)$$ tried to manipulate the equation few times and failed while trying to reach from left side to right side.

would be glad if someone can show a hint how to continue / solve it . Thanks

Tal Ouzan
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1 Answers1

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

  • $\ds{\sum_{k = 1}^{n}x^{2k - 1} = {x^{2n + 1} - x \over x^{2} - 1}}$.
  • Derive, both members, respect of $\ds{x}$: $$ \sum_{k = 1}^{n}x^{2k - 2}\,\pars{2k - 1} = {\pars{2n - 1}x^{2n + 2} - \pars{2n + 1}x^{2n} + x^{2} + 1 \over \pars{x^{2} - 1}^{2}} $$
  • Set $\ds{x = \ic}$: $$ \sum_{k = 1}^{n}\pars{-1}^{k + 1}\,\pars{2k - 1} = \bbx{\pars{-1}^{n + 1}\,\,n} \\ $$
Felix Marin
  • 89,464