Show that $\displaystyle \sum (-1)^{r+1} r^2 $ = -n(2n+1)
summation is from r=1 to 2n . sry i dont know how to type
Thx
hi
\sum^{2n}_{r=1}(-1)^{r+1} r^2 will generate
$\displaystyle \sum^{2n}_{r=1}(-1)^{r+1} r^2$
$\displaystyle =1-4+9-16+25-36+...$
$\displaystyle =1^2-2^2+3^2-4^2+...+(2n-1)^2-(2n)^2$
$\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n-1)$
$\displaystyle =-3-7-...-(4n-1)$
which is an AP with first term -3 , last term -(4n-1)
$\displaystyle S_n=\frac{n}{2}(-3-(4n-1))$
simplify this and you are done.