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:)

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- May 22nd 2010, 10:43 PMhelloyingSummation
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:) - May 23rd 2010, 04:25 AMmathaddict
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. - May 23rd 2010, 07:10 AMoldguynewstudent
- May 24th 2010, 01:06 AMhelloying
- May 27th 2010, 11:38 PMmathaddict