∑_n=1^n1/log_2^n (a) = ∑_n=1^n log_a 2^n = ∑_n=1^nn log_a 2. = log_a 2. ∑_n=1^nn =log_a 2.n(n+1)/2 =n(n+1)/2log_a 2 ...

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Fundamental Laws of Logarithms

∑n=1n1/log2n ...

Question

$\sum_{n = 1}^{n} \frac{1}{l o g_{2^{n}} (a)} =$

\frac{n (n + 1)}{2} l o g_{a} 2

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\frac{n (n + 1)}{2} l o g_{2} a

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\frac{(n + 1)^{2} n^{2}}{4} l o g_{2} a

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N o n e o f t h e s e

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Solution

The correct option is A $\frac{n (n + 1)}{2} l o g_{a} 2$
$\sum_{n = 1}^{n} \frac{1}{l o g_{2^{n}} (a)}$
$= \sum_{n = 1}^{n} l o g_{a} 2^{n}$
$= \sum_{n = 1}^{n} n l o g_{a} 2.$
$= l o g_{a} 2. \sum_{n = 1}^{n} n$
$= l o g_{a} 2. \frac{n (n + 1)}{2}$
$= \frac{n (n + 1)}{2} l o g_{a} 2$

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