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Question

Question: \[\sum_{n = 1}^{n}\frac{1}{\log_{2^{n}}(a)} =\]...

∑n=1n1log⁡2n(a)=\sum_{n = 1}^{n}\frac{1}{\log_{2^{n}}(a)} =

A

n(n+1)2log⁡a2\frac{n(n + 1)}{2}\log_{a}2

B

n(n+1)2log⁡2a\frac{n(n + 1)}{2}\log_{2}a

C

(n+1)2n24log⁡2a\frac{(n + 1)^{2}n^{2}}{4}\log_{2}a

D

None of these

Answer

n(n+1)2log⁡a2\frac{n(n + 1)}{2}\log_{a}2

Explanation

Solution

∑n=1n1log⁡2n(a)=∑n=1nlog⁡a2n\sum_{n = 1}^{n}\frac{1}{\log_{2^{n}}(a)} = \sum_{n = 1}^{n}{\log_{a}2^{n}}=x=1x = 1

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