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Basic of Logarithms
medium
$\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
એકપણ નહી.
Solution
(a) $\sum\limits_{n = 1}^n {{1 \over {{{\log }_{{2^n}}}(a)}}} = \sum\limits_{n = 1}^n {{{\log }_a}{2^n}} = x = 1$
$= {\log _a}2.{{n(n + 1)} \over 2} = {{n(n + 1)} \over 2}{\log _a}2$.
Standard 11
Mathematics
