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