The correct option is A 1/2log_e2 We have identity x + #160;13x^3 + 15x^5 +.... = 12 log(1+x1 x) We have, #160; 1/3+1/3(1/3)^3+1/5(1/3)^5+…. ...

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Proof by mathematical induction

1/3+1/31/33+1...

Question

$\frac{1}{3} + \frac{1}{3} (\frac{1}{3})^{3} + \frac{1}{5} (\frac{1}{3})^{5} + \dots . =$

\frac{1}{2} {log}_{e} 2

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2 {log}_{e} 2

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{log}_{e} 2

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{log}_{e} 3

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Solution

The correct option is A $\frac{1}{2} {log}_{e} 2$
We have identity

$x + \frac{1}{3} x^{3} + \frac{1}{5} x^{5} + . . . . = \frac{1}{2} l o g (\frac{1 + x}{1 - x})$

We have, $\frac{1}{3} + \frac{1}{3} (\frac{1}{3})^{3} + \frac{1}{5} (\frac{1}{3})^{5} + \dots .$

Substitute $x = \frac{1}{3}$ and solve

$= \frac{1}{2} l o g (\frac{1 + \frac{1}{3}}{1 - \frac{1}{3}})$

$= \frac{1}{2} {log}_{e} 2$

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Similar questions

Q. Find the sum of the following series

(1 + 2) l o g_{e} 2 + \frac{1 + 2^{2}}{⌊ 2} (l o g_{e} 2)^{2} + \frac{1 + 2^{3}}{⌊ 3} (l o g_{e} 2)^{3} + . . . .

Q. Match the elements of List I with List II:

List-I	List-II
A) $\int_{0}^{\frac{π}{2}} log sin x d x =$	1) $4 π$
B) $\int_{0}^{\frac{π}{2}} log tan x d x =$	2) $- \frac{π}{2} {log}_{e} 2$
C) $\int_{0}^{π} x log sin x d x =$	3) $- \frac{π^{2}}{2} {log}_{e} 2$
D) $\int_{- π}^{π} (x^{3} + x cos x + {tan}^{5} x + 2) d x =$	4) $0$

Q. If

a = 1, b = {log}_{e} 2, c = {log}_{e} 3

, then which condition is true?

Q. The value of the integral

1 \int 0 x {cot}^{- 1} (1 - x^{2} + x^{4}) d x

is :

Find the sum of infinite series $\frac{1 + 3}{1!} (l o g_{e} 3) + \frac{1 + 3^{2}}{2!} (l o g_{e} 3)^{2} + \frac{1 + 3^{3}}{3!} (l o g_{e} 3)^{3} + . . .$

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13+13(13)3+15(13)5+….=

The correct option is A 12loge2We have identityx+13x3+15x5+....=12log(1+x1−x)We have, 13+13(13)3+15(13)5+….Substitute x=13 and solve =12log(1+131−13)=12loge2

$\frac{1}{3} + \frac{1}{3} (\frac{1}{3})^{3} + \frac{1}{5} (\frac{1}{3})^{5} + \dots . =$