First slide

Special Series in Sequences and Series

Question

Sum to n terms of the series
$\tan^{- 1} (\frac{1}{3}) + \tan^{- 1} (\frac{1}{7}) + \tan^{- 1} (\frac{1}{13}) + \dots$ is

Moderate

A

$\tan^{- 1} (\frac{n}{n + 2})$
B

$\tan^{- 1} (\frac{2 n - 1}{2 n + 1})$
C

$\tan^{- 1} (\frac{1}{3 n})$
D

$\tan^{- 1} (\frac{1}{2 n})$

Solution

$a_{r} = \tan^{- 1} (\frac{1}{1 + r (r + 1)}) = \tan^{- 1} (\frac{r + 1 - r}{1 + (r + 1) r})$
$= \tan^{- 1} (r + 1) - \tan^{- 1} (r)$
Thus , $S_{n} = \sum_{r = 1}^{n} a_{r} = \tan^{- 1} (n + 1) - \tan^{- 1} (1)$
$= \tan^{- 1} (\frac{n + 1 - 1}{1 + (n + 1) (1)}) = \tan^{- 1} (\frac{n}{n + 2})$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App