Induction/Alternating sum of squares/Exercise/Solution

Initial step. For ${}n=1$ we have on the left hand side only the summand for ${}k=2$, which is

${}(-1)^{0}1^{2}=1\,.$

On the right hand side we also have

${}(-1)^{2}{\frac {1\cdot 2}{2}}=1\,.$

Induction step. Suppose that the statement ist already proven for ${}n$, we have to show that it is also true for ${}n+1$. We have

Failed to parse (unknown function "\begin{align}"): {\displaystyle {{}} \begin{align} \sum_{k <table class="metadata plainlinks ambox ambox-notice" style=""> <tr> <td class="mbox-image"><div style="width: 52px;"> [[File:Wikiversity logo 2017.svg|50px|link=]]</div></td> <td class="mbox-text" style=""> '''[[m:Soft redirect|Soft redirect]]'''<br />This page can be found at <span id="SoftRedirect">[[mw:Help:Magic words#Other]]</span>. </td> </tr> </table>[[Category:Wikiversity soft redirects|Induction/Alternating sum of squares/Exercise/Solution]] __NOINDEX__ 1}^{n+1} (-1)^{k-1} k^2 & = \sum_{k <table class="metadata plainlinks ambox ambox-notice" style=""> <tr> <td class="mbox-image"><div style="width: 52px;"> [[File:Wikiversity logo 2017.svg|50px|link=]]</div></td> <td class="mbox-text" style=""> '''[[m:Soft redirect|Soft redirect]]'''<br />This page can be found at <span id="SoftRedirect">[[mw:Help:Magic words#Other]]</span>. </td> </tr> </table>[[Category:Wikiversity soft redirects|Induction/Alternating sum of squares/Exercise/Solution]] __NOINDEX__ 1}^{n} (-1)^{k-1} k^2 + (-1)^n (n+1)^2 \\ & = (-1)^{n+1} \frac{ n(n+1) }{ 2 } + (-1)^{n+2} (n+1)(n+1) \\ & = (-1)^{n+2} \frac{ -n(n+1) }{ 2 } + \frac{ (-1)^{n+2} 2 (n+1)(n+1) }{ 2 } \\ & = (-1)^{n+2} \frac{ -n(n+1)+ 2(n+1)(n+1) }{ 2 } \\ & = (-1)^{n+2} \frac{ (n+1) (-n+2n+2) }{ 2 } \\ & = (-1)^{n+2} \frac{ (n+1) (n+2) }{ 2 } . \end{align} }

Thus the statement holds for all ${}n$.

To the exercise