If A = ∑ _ n = 1 ^ ∞ 1 (3 + (- 1) ^ n ) ^ n and B = ∑ _ n = 1 ^ ∞ (- 1) ^ n (3 + (- 1) ^ n ) ^ n , then (A)/(B) is equal to

Correct answer: C. - (11)/(9) Given, A = ∑ _ n = 1 ^ ∞ 1 (3 + (- 1) ^ n ) ^ n = (1)/(2) + 1 4 ^ 2 + 1 2 ^ 3 + 1 4 ^ 4 + ⋯ = ((1)/(2) + 1 2 ^ 3 + ⋯) + ( 1 4 ^ 2 + 1 4 ^ 4 + ⋯) = ((1)/(2))/(1 - (1)/(4)) + ((1)/(16))/(1 - (1)/(16)) = (2)/(3) + (1)/(15) = (11)/(15) B = ∑ _ n = 1 ^ ∞ (- 1) ^ n (3 + (- 1) ^ n ) ^ n = - (1)/(2) + 1 4 ^ 2 - 1 2 ^ 3 + 1 4 ^ 4 - ⋯ = - ((1)/(2) + 1 2 ^ 3 + ⋯) + ( 1 4 ^ 2 + 1 4 ^ 4 + ⋯) = - ((1)/(2))/(1 - (1)/(4)) + ((1)/(16))/(1 - (1)/(16)) = - (2)/(3) + (1)/(15) = - (9)/(15) Hence, (A)/(B) = - (11)/(9)

JEE Main 202226 JuneEvening Shift

Exam: JEE Main
Year: 2022
Date: 26 June
Shift: Evening Shift
Subject: Mathematics
Chapter: Sequences and Series

JEE Main 2022 · 26 June · Evening Shift · Mathematics · Sequences and Series

If A = ∑ _ n = 1 ^ ∞ 1 (3 + (- 1) ^ n ) ^ n and B = ∑ _ n = 1 ^ ∞...

A = \mathop{∑}_{n = 1}^{\infty}\dfrac{1}{{\left(3 + {\left(- 1\right)}^{n}\right)}^{n}}

and

B = \mathop{∑}_{n = 1}^{\infty}\dfrac{{\left(- 1\right)}^{n}}{{\left(3 + {\left(- 1\right)}^{n}\right)}^{n}}

, then

\dfrac{A}{B}

is equal to

Options

A $\dfrac{11}{9}$
B $1$
C $- \dfrac{11}{9}$
D $- \dfrac{11}{3}$

Correct answer

C. $- \dfrac{11}{9}$

Step-by-step solution

Given, $A = \mathop{∑}_{n = 1}^{\infty}\dfrac{1}{{\left(3 + {\left(- 1\right)}^{n}\right)}^{n}} = \dfrac{1}{2} + \dfrac{1}{{4}^{2}} + \dfrac{1}{{2}^{3}} + \dfrac{1}{{4}^{4}} + ⋯$

$= \left(\dfrac{1}{2} + \dfrac{1}{{2}^{3}} + ⋯\right) + \left(\dfrac{1}{{4}^{2}} + \dfrac{1}{{4}^{4}} + ⋯\right)$ = \dfrac{\dfrac{1}{2}}{1 - \dfrac{1}{4}} + \dfrac{\dfrac{1}{16}}{1 - \dfrac{1}{16}} $$ = \dfrac{2}{3} + \dfrac{1}{15} = \dfrac{11}{15} $$ B = \mathop{∑}_{n = 1}^{\infty}\dfrac{{\left(- 1\right)}^{n}}{{\left(3 + {\left(- 1\right)}^{n}\right)}^{n}} = - \dfrac{1}{2} + \dfrac{1}{{4}^{2}} - \dfrac{1}{{2}^{3}} + \dfrac{1}{{4}^{4}} - ⋯ $$ = - \left(\dfrac{1}{2} + \dfrac{1}{{2}^{3}} + ⋯\right) + \left(\dfrac{1}{{4}^{2}} + \dfrac{1}{{4}^{4}} + ⋯\right) $$ = - \dfrac{\dfrac{1}{2}}{1 - \dfrac{1}{4}} + \dfrac{\dfrac{1}{16}}{1 - \dfrac{1}{16}} = - \dfrac{2}{3} + \dfrac{1}{15} = - \dfrac{9}{15} $Hence,$ \dfrac{A}{B} = - \dfrac{11}{9}$

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