1 + 2r + r^2 + 2r^3 + r^4 + 2r^5 + r^6 ...

What is the sum of the series when it converges?

I know that geometric sequence converges when r is less than 1, and the sum will converge to:
a/(1-r). But I don't know how to find a and r for this sequence.

Question

1 + 2r + r^2 + 2r^3 + r^4 + 2r^5 + r^6 ...

What is the sum of the series when it converges?


I know that geometric sequence converges when r is less than 1, and the sum will converge to:
a/(1-r). But I don't know how to find a and r for this sequence.

hartnn · Answer

a= 1st term =1

hartnn · Answer

r = common ratio = 2nd term/1st term = 3rd term/2nd term = ....
and so on

hartnn · Answer

so whats, 
2r/1 = r^2/2r ...
wait its not geometric...

hartnn · Answer

ohh......we split it into 2 geometric sequences....

hartnn · Answer

1+r^2+r^4+.......
2r+2r^3+.....

hartnn · Answer

the 1st sequence has common ratio of r^2
could u understand how?

anonymous · Answer

How did you know that you had to split it?
And yes I see that the common ratio is r^2 for the first sequence

hartnn · Answer

what about the 2nd sequence ? 

i could just figure out that alternate terms were in GP

hartnn · Answer

GP =geometric progression

anonymous · Answer

is a = r in the second sequence?
so sum for first sequence is: 1/(1-r^2)
second sequence is: 2r/(1-r^2) ?
Then we add these?

hartnn · Answer

absolutely correct!

anonymous · Answer

Thank you :D

hartnn · Answer

welcome ^_^