UPSC NDA 2 2017 Mathematics Solved Paper

Concept:The sum of a series of the form $\sum\limits_{k=1}^{n} k \cdot r^{k}$ is found by multiplying the series by $r$ and subtracting the original series.Explanation:Let $S = 1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \dots + n \cdot 3^{n}$.Set $x = 3$, so $S = 1 \cdot x + 2 \cdot x^{2} + \dots + n \cdot x^{n}$. (Equation 1)Multiply (1) by $x$: $xS = 1 \cdot x^{2} + 2 \cdot x^{3} + \dots + n \cdot x^{n+1}$. (Equation 2)Subtract (2) from (1): $S - xS = (x + x^{2} + \dots + x^{n}) - n x^{n+1}$.Factor left side: $S(1-x) = \frac{x(1-x^{n})}{1-x} - n x^{n+1}$, using the sum of a geometric series.Rewrite with $x-1$: $S(1-x) = -\frac{x(x^{n}-1)}{x-1} - n x^{n+1}$.Solve for $S$: $S = \frac{1}{x-1} \left[\frac{x(x^{n}-1)}{x-1} + n x^{n+1}\right] = \frac{x(x^{n}-1) + n x^{n+1}(x-1)}{(x-1)^{2}}$.Substitute $x=3$ (so $x-1=2$): $S = \frac{3(3^{n}-1) + n \cdot 3^{n+1}(2)}{2^{2}} = \frac{3^{n+1} - 3 + 2n \cdot 3^{n+1}}{4} = \frac{3^{n+1}(2n-1) + 3}{4}$.Compare with the given RHS $\frac{(2n-1)3^{a} + b}{4}$. Matching powers and constant terms gives $a = n+1$ and $b = 3$.Answer:$a = n+1$, $b = 3$, which corresponds to option D.