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The sum to infinity of the series 1+2...

The sum to infinity of the series `1+2/3+6/(3^2)+(10)/(3^3)+(14)/(3^4). . . . . .` is (1) 2 (2) 3 (3) 4 (4) 6

A

2

B

3

C

4

D

6

Text Solution

Verified by Experts

The correct Answer is:
B
Let `S=1+2/3+6/3^(2)+10/3^(3)+14/3^(4)`+….
`therefore1/3S=1/3+2/3^(2)+6/3^(3)+10/3^(4)+….`
subtracting (2) from (1), we get
`S(1-1/3)=1+1/3+4/3^(2)+4/3^(3)+4/3^(4)+…..`
`rArr2/3S=4/3+4/3^(2)(1+1/3+1/3^(2)+…)`
`=4/3+4/3^(2)(1/(1-1/3)=4/3+4/3^(2)3/2=4/3+2/3=6/3`
`rArrS=3`
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