UPSC NDA 2 2017 Mathematics Solved Paper

Concept:The product of powers with the same base is obtained by adding the exponents, which here form an infinite geometric series.Explanation:The given expression is $6^{\frac{1}{2}} \times 6^{\frac{1}{4}} \times 6^{\frac{1}{8}} \times 6^{\frac{1}{16}} \times \cdots$ (infinite terms).Using the rule $a^m \times a^n = a^{m+n}$, the product becomes $6^{\left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots\right)}$.The exponents form an infinite geometric series: first term $a = \frac{1}{2}$, common ratio $r = \frac{1}{2}$.The sum of an infinite geometric series is $S = \frac{a}{1 - r}$ (valid for $|r| < 1$).Thus, $S = \frac{\frac{1}{2}}{1 - \frac{1}{2}} = \frac{\frac{1}{2}}{\frac{1}{2}} = 1$.So the product equals $6^{1} = 6$.Answer:$6$