If A=∑n=1∞13+-1^nn and B=∑n=1∞-1^n3+-1^nn, then A/B is equal to

Correct answer: (C) -11/9. Given, A=∑n=1∞(13+-1^nn=1)/2+(14^2+12^3+14^4+⋯=1)/2+(12^3+⋯+14^2+14^4+⋯=1)/21-1/4+116/(1-116) =2/3+1/15=11/15 B=∑n=1∞-1^n3+-1^nn=-1/2+(14^2-12^3+14^4-⋯=-1)/2+(12^3+⋯+14^2+14^4+⋯=-1)/21-1/4+116/(1-116)=-2/3+1/15=-9/15 Hence, A/B=-11/9

Q33 / 90 · JEE Main 2022 (26 Jun Shift 2) · Mathematics

If $A = \sum_{n = 1}^{\infty} \frac{1}{{(3 + {(- 1)}^{n})}^{n}}$ and $B = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(3 + {(- 1)}^{n})}^{n}}$ , then $\frac{A}{B}$ is equal to

A
$\frac{11}{9}$
B
$1$
Correct answer
C
$- \frac{11}{9}$
D
$- \frac{11}{3}$

Solution

Given, $A = \sum_{n = 1}^{\infty} \frac{1}{{(3 + {(- 1)}^{n})}^{n}} = \frac{1}{2} + \frac{1}{4^{2}} + \frac{1}{2^{3}} + \frac{1}{4^{4}} + \dots$

$= (\frac{1}{2} + \frac{1}{2^{3}} + \dots) + (\frac{1}{4^{2}} + \frac{1}{4^{4}} + \dots)$ $= \frac{\frac{1}{2}}{1 - \frac{1}{4}} + \frac{\frac{1}{16}}{1 - \frac{1}{16}}$

$= \frac{2}{3} + \frac{1}{15} = \frac{11}{15}$

$B = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(3 + {(- 1)}^{n})}^{n}} = - \frac{1}{2} + \frac{1}{4^{2}} - \frac{1}{2^{3}} + \frac{1}{4^{4}} - \dots$

$= - (\frac{1}{2} + \frac{1}{2^{3}} + \dots) + (\frac{1}{4^{2}} + \frac{1}{4^{4}} + \dots)$

$= - \frac{\frac{1}{2}}{1 - \frac{1}{4}} + \frac{\frac{1}{16}}{1 - \frac{1}{16}} = - \frac{2}{3} + \frac{1}{15} = - \frac{9}{15}$

Hence, $\frac{A}{B} = - \frac{11}{9}$

Hence, the correct answer is option: 'C'.

If A=∑n=1∞13+-1nn and B=∑n=1∞-1n3+-1nn, then AB is equal to

Solution

Relevant Questions

If $A = \sum_{n = 1}^{\infty} \frac{1}{{(3 + {(- 1)}^{n})}^{n}}$ and $B = \sum_{n = 1}^{\infty} \frac{{(- 1)}^{n}}{{(3 + {(- 1)}^{n})}^{n}}$ , then $\frac{A}{B}$ is equal to