Evaluate the sum: sumlimits_{k = 1}^n frac{1} | vedclass

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Question

(A) We are given the sum $S = \sum\limits_{k = 1}^n \frac{1}{\log_{2^k}(a)}$.Using the change of base property $\frac{1}{\log_b a} = \log_a b$,we have

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$\frac{n(n + 1)}{2} \log_a 2$

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(A) We are given the sum $S = \sum\limits_{k = 1}^n \frac{1}{\log_{2^k}(a)}$.Using the change of base property $\frac{1}{\log_b a} = \log_a b$,we have $\frac{1}{\log_{2^k}(a)} = \log_a(2^k)$.Thus,$S = \sum\limits_{k = 1}^n \log_a(2^k)$.Using the property $\log_a(x^y) = y \log_a x$,we get $S = \sum\limits_{k = 1}^n k \log_a 2$.Since $\log_a 2$ is a constant,$S = (\log_a 2) \sum\limits_{k = 1}^n k$.Using the sum formula $\sum\limits_{k = 1}^n k = \frac{n(n + 1)}{2}$,we obtain $S = \frac{n(n + 1)}{2} \log_a 2$.

Evaluate the sum: $\sum\limits_{k = 1}^n \frac{1}{\log_{2^k}(a)}$

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