`1/(log_3e)+1/log_3e^2+1/log_3e^4+...=`

Options

A. `log_(e)9`
B. 0
C. 1
D. `log_(e)3`

Answer

A

Explanation

`(1)/(log_(3)e)+(1)/(log_(3)e^(2))+(1)/(log_(3)e^(4))+....`
`=(1)/(log_(3)e)+(1)/(2log_(3)e)+(1)/(4log_(3)e)+...("Since, "log_(a)b^(m)=mlog_(a)b)`
`=log_(e)3+(log_(e)3)/(2)+(log_(2)3)/(4)+....("Since, "log_(a)b=(1)/(log_(b)a))`
`=log_(2)3(1+(1)/(2)+(1)/(4)+...)`
`=log_(2)3((1)/(1-(1)/(2)))(because1,(1)/(2),(1)/(4)...." is G.P. with "a=1,r=(1)/(2))`
`=log_(e)3(2)=2log_(e)3=log_(e)9.`