..
(2/3)^n-1
to infinite or 4
no i am asking till where we have to find infinite or 4
@kanwal32
you mean this?
yes, thats what he is asking. But do u understand the solution given in the old post? u can simply substitute 3/2 by 2/3 to get ur ans...
there is a formula given in the old post on how to calculate the sum....did u see it?
I did, but it's pretty confusing, could you start me out?
u need to understand the 3 terms used in the formula: a, R and n a is the initial term in the series so in this case, it is equal to \[(\frac{ 2 }{ 3})^{1-1} \] =1 ok?
Okay.
R is the common ratio...so what is the ratio between the first and second term in this series? Basically what u get if u divide the second term by the first term? u already know the first term, a, is 1...
nope, first term is 1....what is the second term - when n=2....
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