The sum of the series log42-log82+log162-... is equal to
e2
loge2+1
loge3-2
1-loge2
Explanation for the correct option:
The given series can be written as
log222-log232+log242...=12-13+14...∵logyn(xm)=mnlogy(x)andlogxx=1=--12+13-14...=-log(1+1)-1∵log(1+x)=x-x22+x33...,Putting,x=1=1-log(2)=1-loge2
Hence, the correct answer is option (D).