# Summation

• May 22nd 2010, 11:43 PM
helloying
Summation
Show that $\sum (-1)^{r+1} r^2$ = -n(2n+1)

summation is from r=1 to 2n . sry i dont know how to type

Thx:)
• May 23rd 2010, 05:25 AM
Quote:

Originally Posted by helloying
Show that $\sum (-1)^{r+1} r^2$ = -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

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

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

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

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

$=-3-7-...-(4n-1)$

which is an AP with first term -3 , last term -(4n-1)

$S_n=\frac{n}{2}(-3-(4n-1))$

simplify this and you are done.
• May 23rd 2010, 08:10 AM
oldguynewstudent
Quote:

hi

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

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

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

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

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

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

which is an AP with first term -3 , last term -(4n-1)

$S_n=\frac{n}{2}(-3-(4n-1))$

simplify this and you are done.

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

When I do this I come up with $=1^2-2^2+3^2-4^2+...-(2n)^2 +(2n-1)^2$, because $(-1)^{2n+1} {2n}^2$ = $-{2n}^2$.
• May 24th 2010, 02:06 AM
helloying
Quote:

Originally Posted by oldguynewstudent

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

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

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

it shd be -3-7 and not -3-4
• May 28th 2010, 12:38 AM
yep i realise i dont get it. i agree with oldguynewstudnet.and also in the part $