JEE Main 2022 (Online) 26th June Evening ShiftQuestion 1
If $A=\sum_{n=1}^{\infty}\frac{1}{\left(3+(-1)^{n}\right)^{n}}$ and $B=\sum_{n=1}^{\infty}\frac{(-1)^{n}}{\left(3+(-1)^{n}\right)^{n}}$, then $\frac{A}{B}$ is equal to :
Sequence and Series · Geometric Progression
Solution
Formulas used. (1) $3+(-1)^n=2$ for odd $n$, and $4$ for even $n$(2) Sum of an infinite G.P.: $\frac{\text{first}}{1-r}$Step 1 - Odd $n$. $$\sum_{n\ \text{odd}}\frac1{2^n}=\frac12+\frac18+\cdots=\frac{1/2}{1-1/4}=\frac23 .$$ Step 2 - Even $n$. $$\sum_{n\ \text{even}}\frac1{4^n}=\frac1{16}+\frac1{256}+\cdots=\frac{1/16}{1-1/16}=\frac1{15}.$$ Step 3 - $A$ and $B$. In $B$, odd $n$ carry a minus sign: $$A=\frac23+\frac1{15}=\frac{11}{15},\qquad B=-\frac23+\frac1{15}=-\frac35 .$$ Step 4. $$\frac AB=\frac{11/15}{-9/15}=-\frac{11}{9}.$$ Answer: (C)