$\sum\limits_{n = 1}^n {{1 \over {{{\log }_{{2^n}}}(a)}}} = $
$\sum\limits_{n = 1}^n {{1 \over {{{\log }_{{2^n}}}(a)}}} = $
- A
${{n(n + 1)} \over 2}{\log _a}2$
- B
${{n(n + 1)} \over 2}{\log _2}a$
- C
${{{{(n + 1)}^2}{n^2}} \over 4}{\log _2}a$
- D
None of these
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