The sum to infinity of the series `1+2/3+6/3^2+14/3^4+...is `

Options

A. 2
B. 3
C. 4
D. 6

Answer

B

Explanation

Let S be the sum of the given series i.e.,
`S=1+(2)/(3)+(6)/(3^(2))+(10)/(3^(3))+(14)/(3^(4))+ . . . . . "to "oo`
`rArr" "(S-1)=(2)/(3)+(6)/(3^(2))+(10)/(3^(3))+(14)/(3^(4))+ . . . . ."to "oo` . . .(i)
`rArr" "(S-1)xx(1)/(3)=(2)/(3^(2))+(6)/(3^(3))+(10)/(3^(4))+ . . . "to "oo` . . . .(ii)
Subtracting (ii) from (i) we get
`(2)/(3)(S-1)=(2)/(3)+(3^(4/2))/(1-(1)/(3))rArr(2)/(3)(S-1)=(2)/(3)(S-1)=(2)/(3)+(2)/(3)rArrS=3`
ALITER From (i), we observe that 3(S-1) is an infinite arithmetic- gemoetric series with a=2, d=4 and `r=(1)/(3)`.
`:." "3(S-1)=(1)/(1-(1)/(3))+(4xx(1)/(3))/((1-(1)/(3))^(2))=3+3rArrS=3`.