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Let S=4/19+44/(19)^2+444/(19)^3+...oo th...

Let S=`4/19+44/(19)^2+444/(19)^3+...oo` then find the value of S

A

`(38)/(81)`

B

`(4)/(19)`

C

`(36)/(871)`

D

`(4)/(9)`

Text Solution

Verified by Experts

The correct Answer is:
A
The sum of the series ………..
`S = (4)/(19)+(44)/(19^(2))+………… oo`
`(S)/(19) = (4)/(19^(2))+(44)/(19^(3))+………….oo`
`(18S)/(19)= (4)/(19)[1+(10)/(19)+(10^(2))/(19^(2))+………..oo]`
`s = (38)/(81)`
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