If the sum to infinity of the series `1+2r+3r^2+4r^3+` is 9/4, then value of `r` is `1//2` b. `1//3` c. `1//4` d. none of these

Options

A. `1//2`
B. `1//3`
C. `1//4`
D. none of these

Answer

B

Explanation

`S=1+2r+3r^(2)+4r^(3)+..`
`thereforerS=r+2r^(2)+3r^(3)+4r^(4)+..`
`rArr(1-r)S=1+r+r^(2)+r^(3)+..`
`=1/(1-r)`
`rArrS=1/((1-)^(2))`
Given `S=9/4`
`rArr1/((1-r)^(2))=9/4`
`rArr1-r=pm2/3`
`rArrr=1/3or5/3`
Hence, r=1/3 as `0ltabsrlt1`