श्रेणी `1/3+5/9+19/27+65/81+...`के `n` पदों का योग है

Options

A. `2^n-1`
B. `3^n-1`
C. `n-((3^n-2^n))/2^n`
D. `n-(2-(3^n-2^n))/3^n`

Answer

D

Explanation

माना
`S+(1-2/3)+(1-4/9)+(1-9/27)+...`
`=n-[2/3+(2/3)^2+(2/3)^3+...+n"पद"]`
`=n-2/3([1-(2/3)^n])/(1-2/3)=n-(2(3^n-2^n))/3^n`