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

${}{\begin{aligned}\sum _{k=1}^{n+1}(-1)^{k-1}k^{2}&=\sum _{k=1}^{n}(-1)^{k-1}k^{2}+(-1)^{n}(n+1)^{2}\\&=(-1)^{n+1}{\frac {n(n+1)}{2}}+(-1)^{n+2}(n+1)(n+1)\\&=(-1)^{n+2}{\frac {-n(n+1)}{2}}+{\frac {(-1)^{n+2}2(n+1)(n+1)}{2}}\\&=(-1)^{n+2}{\frac {-n(n+1)+2(n+1)(n+1)}{2}}\\&=(-1)^{n+2}{\frac {(n+1)(-n+2n+2)}{2}}\\&=(-1)^{n+2}{\frac {(n+1)(n+2)}{2}}.\end{aligned}}$

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

To the exercise