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

*

*$\ds{\sum_{k = 1}^{n}x^{2k - 1} =
{x^{2n + 1} - x \over x^{2} - 1}}$.

*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 $\ds{x = \ic}$:
$$
\sum_{k = 1}^{n}\pars{-1}^{k + 1}\,\pars{2k - 1} =
\bbx{\pars{-1}^{n + 1}\,\,n} \\
$$
