i have a simple question.

can someone please tell me how to get there?

I just dont get it.

http://i709.photobucket.com/albums/w...ntitled5-1.jpg

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- May 9th 2009, 09:16 AMkrispiekreamsimple solution?
i have a simple question.

can someone please tell me how to get there?

I just dont get it.

http://i709.photobucket.com/albums/w...ntitled5-1.jpg - May 9th 2009, 09:27 AMkrispiekream
Also,

What can i do in this case?

c(j)c(j + 2)??? - May 9th 2009, 10:34 AMHallsofIvy
$\displaystyle \sum_{j= 0}^\infty \left(0.9\right)^{2j}= \sum_{j=0}^\infty \left(0.9^2\right)^j= \sum_{j= 0}^\infty \left(0.81\right)^j$

is a geometric series with a= 1 and r= 0.81. Use the formula for the sum of a geometric series:$\displaystyle \sum_{j=0}^\infty ar^j= \frac{a}{1- r}$ - May 9th 2009, 10:43 AMkrispiekream
thank you..i needed that..

- May 9th 2009, 11:56 AMkrispiekream
I understand the first step now..

what about the second step?

where is 0.9* geometric series come from?

http://i709.photobucket.com/albums/w...ntitled5-2.jpg

is it because c(n)=(0.9)^n

c(0) for n=0, therefore

c(0)=1

c(1)=0.9

c(2)=0.81?