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1+2/3+6/(3^2)+10/(3^3)+14/(3^4)+...oo=...

`1+2/3+6/(3^2)+10/(3^3)+14/(3^4)+...oo=`

A

6

B

2

C

3

D

4

Text Solution

Verified by Experts

The correct Answer is:
C
Let `S=1+(2)/(3)+(6)/(3^(2))+(10)/(3^(3))+(14)/(3^(4))+"......"" " "……..(i)`
`:. (1)/(3)S=(1)/(3)+(4)/(3^(2))+(6)/(3^(3))+(10)/(3^(4))+"......"" " "........(ii)"`
On subtracting Eq. (ii) from Eq.(i), we get
`(2)/(3)S=1+(1)/(3)+(4)/(3^(2))+(4)/(3^(3))+(4)/(3^(4))+"......"`
`=(4)/(3)+4((1)/(3^(2))+(1)/(3^(3))+(1)/(3^(4))+"......")=(4)/(3){((1)/(3^(2)))/(1-(1)/(3))}=(4)/(3)+(2)/(3)=2`
`:. S=3`.
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