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(2/3)^n-1

to infinite or 4

no i am asking till where we have to find infinite or 4

you mean this?

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?

Okay.

nope, first term is 1....what is the second term - when n=2....

|dw:1434612184951:dw|

no, ignore the part that I scribbled out...whatever is left, substitute n=2 and what will u get?

I think it is 3

\[(\frac{ 2 }{ 3 })^{(2-1)}\] = 2/3
So that is R in the formula ok?

(2/3) (2 - 1) is 2/3. now what? @sdfgsdfgs

What is the formula to get the sum from the old post?

oh okay..

a=1
R=2/3
n=4
Calculate S which is ur ans.

Put in the values we found for a, R and n....

where would a go?

Correct! It is
\[S = \frac{ (1-(\frac{ 2 }{ 3 })^4 )}{ 1-\frac{ 2 }{ 3 } }\]

im right there and i kind of understand it a bit now..

try it again - 65/16 is not the right ans.

okay one sec..

sorry i need to go now....65/27 is the right ans. good luck!