UPSC NDA II 2015 Mathematics Paper

Concept:The infinite product can be expressed as $6$ raised to the sum of a series of exponents, which forms an arithmetico-geometric progression (A.G.P.).Explanation:Rewrite the product as $S = 6^{\frac{1}{2}} \times 6^{\frac{2}{4}} \times 6^{\frac{3}{8}} \times 6^{\frac{4}{16}} \times \cdots$.The exponents follow the pattern $\frac{n}{2^n}$ for $n = 1, 2, 3, \dots$.Thus, $S = 6^{\sum_{n=1}^{\infty} \frac{n}{2^n}}$.Let $x = \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \cdots$ — (1).Multiply both sides by $\frac{1}{2}$: $\frac{x}{2} = \frac{1}{4} + \frac{2}{8} + \frac{3}{16} + \frac{4}{32} + \cdots$ — (2).Subtract (2) from (1): $x - \frac{x}{2} = \frac{1}{2} + \left(\frac{2}{4}-\frac{1}{4}\right) + \left(\frac{3}{8}-\frac{2}{8}\right) + \left(\frac{4}{16}-\frac{3}{16}\right) + \cdots$.This simplifies to $\frac{x}{2} = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots$.The right side is a geometric series with first term $a = \frac{1}{2}$ and common ratio $r = \frac{1}{2}$.Its sum is $\frac{a}{1-r} = \frac{1/2}{1-1/2} = 1$.Hence, $\frac{x}{2}=1$ ⇒ $x=2$.Therefore, $S = 6^{2} = 36$.Answer:$36$ (Option B).