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

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

Mathematics - Sequence & Series Question with Solution | TestHub

MathematicsSequence & SeriesG.P.Medium2 minPYQ_2022

MathematicsMediumstatement

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

Options:

Answer:

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

Stream:JEESubject:MathematicsTopic:Sequence & SeriesSubtopic:G.P.

⏱ 2mℹ️ Source: PYQ_2022

Doubts & Discussion

Loading discussions...