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

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

Thus the statement holds for all $n$.

To the exercise