sumlimits_{n = 1}^n {{1 over {{{log }_{{2^n}} | vedclass

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Question

(a) $\sum\limits_{n = 1}^n {{1 \over {{{\log }_{{2^n}}}(a)}}} = \sum\limits_{n = 1}^n {{{\log }_a}{2^n}} = x = 1$

$= {\log _a}2.{{n(n + 1)} \over 2

Vedclass · Accepted Answer

${{n(n + 1)} \over 2}{\log _a}2$

Vedclass · Answer

(a) $\sum\limits_{n = 1}^n {{1 \over {{{\log }_{{2^n}}}(a)}}} = \sum\limits_{n = 1}^n {{{\log }_a}{2^n}} = x = 1$

$= {\log _a}2.{{n(n + 1)} \over 2} = {{n(n + 1)} \over 2}{\log _a}2$.

$\sum\limits_{n = 1}^n {{1 \over {{{\log }_{{2^n}}}(a)}}} = $

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