simple solution?

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May 9th 2009, 09:16 AM
krispiekream

simple 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 AM
krispiekream

Also,
What can i do in this case?
c(j)c(j + 2)???
May 9th 2009, 10:34 AM
HallsofIvy

Quote:

Originally Posted by krispiekream

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

$\sum_{j= 0}^\infty \left(0.9\right)^{2j}= \sum_{j=0}^\infty \left(0.9^2\right)^j= \sum_{j= 0}^\infty \left(0.81\right)^j$
is a geometric series with a= 1 and r= 0.81. Use the formula for the sum of a geometric series: $\sum_{j=0}^\infty ar^j= \frac{a}{1- r}$
May 9th 2009, 10:43 AM
krispiekream

thank you..i needed that..
May 9th 2009, 11:56 AM
krispiekream

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?

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