_ n = 1 ^n 1 _ 2^n (a) = - Vidyadip

MCQ

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

✓

Answer

Correct option: A.

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

a
(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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