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Question: The sum of the series \(\frac{1}{\log_{2}a} + \frac{1}{\log_{4}a} + \frac{1}{\log_{8}a} + .....\)upt...

The sum of the series 1log⁡2a+1log⁡4a+1log⁡8a+.....\frac{1}{\log_{2}a} + \frac{1}{\log_{4}a} + \frac{1}{\log_{8}a} + .....upto n terms is

A

n(n+1)2\frac{n(n + 1)}{2}loga2

B

n2\frac{n}{2} loga2

C

(n+1)2\frac{(n + 1)}{2}loga2

D

n(n+1)(2n+1)6\frac{n(n + 1)(2n + 1)}{6}loga2

Answer

n(n+1)2\frac{n(n + 1)}{2}loga2

Explanation

Solution

The given expression can be written as

loga2 + loga4 + loga8 + .... upto n terms

= loga2 + 2 loga2 + 3 loga2 + .... n loga2

= loga2 [1+ 2 + ... n] = n(n+1)2log⁡a2\frac{n(n + 1)}{2}\log_{a}2