# Summation

• May 22nd 2010, 10:43 PM
helloying
Summation
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 AM
Quote:

Originally Posted by helloying
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.
• May 23rd 2010, 07:10 AM
oldguynewstudent
Quote:

hi

\sum^{2n}_{r=1}(-1)^{r+1} r^2 will generate

$\displaystyle \sum^{2n}_{r=1}(-1)^{r+1} r^2$

$\displaystyle =1-4+9-16+25-36+...$

$\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$

$\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)$

$\displaystyle =-3-4-...-(4n-1)$

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.

How did you get $\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$?

When I do this I come up with $\displaystyle =1^2-2^2+3^2-4^2+...-(2n)^2 +(2n-1)^2$, because $\displaystyle (-1)^{2n+1} {2n}^2$ = $\displaystyle -{2n}^2$.
• May 24th 2010, 01:06 AM
helloying
Quote:

Originally Posted by oldguynewstudent

How did you get $\displaystyle =1^2-2^2+3^2-4^2+...+(2n)^2-(2n-1)^2$?

When I do this I come up with $\displaystyle =1^2-2^2+3^2-4^2+...-(2n)^2 +(2n-1)^2$, because $\displaystyle (-1)^{2n+1} {2n}^2$ = $\displaystyle -{2n}^2$.

yep i realise i dont get it. i agree with oldguynewstudnet.and also in the part $\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)$

it shd be -3-7 and not -3-4
• May 27th 2010, 11:38 PM
Quote:

Originally Posted by helloying
yep i realise i dont get it. i agree with oldguynewstudnet.and also in the part $\displaystyle =(1+2)(1-2)+(3+4)(3-4)+...+(2n+2n-1)(2n-2n+1)$

it shd be -3-7 and not -3-4

yup , some algebraic error here and there . Edited .