Evaluate the sum: $\sum\limits_{k = 1}^n {\frac{1}{{{{\log }_{{2^k}}}(a)}}}$
- A$\frac{{n(n + 1)}}{2}{\log _a}2$
- B$\frac{{n(n + 1)}}{2}{\log _2}a$
- C$\frac{{{{(n + 1)}^2}{n^2}}}{4}{\log _2}a$
- DNone of these
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