The sum to infinite terms of the series (1/1-(1)4)+(1/(1+3)-(1)4)+(1/(1+3+5)-(1)4)+ ldots ldots . . ∞ is

Question

The sum to infinite terms of the series (1/1-(1)4)+(1/(1+3)-(1)4)+(1/(1+3+5)-(1)4)+ ldots ldots . . ∞ is (A) 1 (B) 2 (C) (1/2) (D) 3

Tardigrade Admin · Accepted Answer

t n =(1/1+3+5+7+ ldots .+(2 n -1)-(1)4) =(1/ n 2-(1)4)=(4/4 n 2-1)=(4/(2 n -1)(2 n +1))=2 (1/2 n -1)-(1/2 n +1) ∴ S n = Sigma t n =2 (1/1)-(1/2 n +1) =(4 n /2 n +1) .

Q. The sum to infinite terms of the series $\frac{1}{1 - \frac{1}{4}} + \frac{1}{( 1 + 3 ) - \frac{1}{4}} + \frac{1}{( 1 + 3 + 5 ) - \frac{1}{4}} + \dots\dots ..\infty$ is

1

2

$\frac{1}{2}$

3

$t_{n} = \frac{1}{1 + 3 + 5 + 7 + \dots . + ( 2 n - 1 ) - \frac{1}{4}}$
$= \frac{1}{n ^{2} - \frac{1}{4}} = \frac{4}{4 n ^{2} - 1} = \frac{4}{( 2 n - 1 ) ( 2 n + 1 )} = 2 {\frac{1}{2 n - 1} - \frac{1}{2 n + 1}}$
$∴ S_{n} = Σ t_{n} = 2 {\frac{1}{1} - \frac{1}{2 n + 1}} = \frac{4 n}{2 n + 1} .$

Q. The sum to infinite terms of the series 1−41​1​+(1+3)−41​1​+(1+3+5)−41​1​+……..∞ is

1

2

21​

3

tn​=1+3+5+7+….+(2n−1)−41​1​ =n2−41​1​=4n2−14​=(2n−1)(2n+1)4​=2{2n−11​−2n+11​} ∴Sn​=Σtn​=2{11​−2n+11​}=2n+14n​.

Q. The sum to infinite terms of the series $\frac{1}{1 - \frac{1}{4}} + \frac{1}{( 1 + 3 ) - \frac{1}{4}} + \frac{1}{( 1 + 3 + 5 ) - \frac{1}{4}} + \dots\dots ..\infty$ is

$\frac{1}{2}$

$t_{n} = \frac{1}{1 + 3 + 5 + 7 + \dots . + ( 2 n - 1 ) - \frac{1}{4}}$
$= \frac{1}{n ^{2} - \frac{1}{4}} = \frac{4}{4 n ^{2} - 1} = \frac{4}{( 2 n - 1 ) ( 2 n + 1 )} = 2 {\frac{1}{2 n - 1} - \frac{1}{2 n + 1}}$
$∴ S_{n} = Σ t_{n} = 2 {\frac{1}{1} - \frac{1}{2 n + 1}} = \frac{4 n}{2 n + 1} .$