In a geometric series , the first term is a and common ratio is r. If `S_n` denotes the sum of the n terms and `U_n=Sigma_(n=1)^(n) S_n` , then `rS_n+(1-r)U_n` equals

Options

A. 0
B. n
C. na
D. nar

Answer

C

Explanation

`S_(n)=(a(r^(n)-1))/(r-1)`
`rArrU_(n)=sum_(n=1)^(n)(a(r^(n)-1))/(r-1)=a/(r-1)sum_(n=1)^(n)(r^(n)-1)`
`=a/(r-1){r+r^(2)+…+r^(n)-n}`
`a/(r-1){(r(r^(n)-1))/(r-1)-n}`
`rArr(r-1)U_(n)=(ar(r^(n)-1))/(r-1)-an`
`rArr(r-1)U_(n)=rS_(n)-an`
`rArrrS_(n)+(1-r)U_(n)=an`