Question

The sum of the series log4 2 - log8 2 + log16 2... is

(A) e2

(B) loge 2

(C) loge3 - 2

(D) 1 - loge2

Answer 1

Correct option**(D) 1- loge 2**

Since logyn xm = m/n logyx and logx x = 1, therefore

• = 1/2 - 1/3 + 1/4 - 1/5 + ....

Also, log (1 + x) = x = x2/2 + x3 /3- x2/2 + ....

Putting x = 1 we have, * = 1 - loge2.