Find the sum of the series: tan^{-1} left( fr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The general term of the series is $T_k = 	an^{-1} \left( \frac{1}{1 + k(k + 1)} ight)$.We can rewrite the argument as $\frac{(k + 1) - k}{1 + k

Vedclass · Accepted Answer

$	an^{-1} \left( \frac{n}{n + 2} ight)$

Vedclass · Answer

(A) The general term of the series is $T_k = 	an^{-1} \left( \frac{1}{1 + k(k + 1)} ight)$.We can rewrite the argument as $\frac{(k + 1) - k}{1 + k(k + 1)}$.Using the identity $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x - y}{1 + xy} ight)$, we get $T_k = 	an^{-1}(k + 1) - 	an^{-1}(k)$.The sum $S_n = \sum_{k=1}^{n} (	an^{-1}(k + 1) - 	an^{-1}(k))$.This is a telescoping series: $S_n = (	an^{-1} 2 - 	an^{-1} 1) + (	an^{-1} 3 - 	an^{-1} 2) + \dots + (	an^{-1}(n + 1) - 	an^{-1} n)$.All intermediate terms cancel out, leaving $S_n = 	an^{-1}(n + 1) - 	an^{-1}(1)$.Using the identity again, $S_n = 	an^{-1} \left( \frac{(n + 1) - 1}{1 + (n + 1)(1)} ight) = 	an^{-1} \left( \frac{n}{n + 2} ight)$.

Find the sum of the series: $\tan^{-1} \left( \frac{1}{1 + 1 \times 2} \right) + \tan^{-1} \left( \frac{1}{1 + 2 \times 3} \right) + \dots + \tan^{-1} \left( \frac{1}{1 + n(n + 1)} \right) =$

Similar Questions

If $\sin^{-1} x + \cos^{-1} y = \frac{2\pi}{5}$,then $\cos^{-1} x + \sin^{-1} y$ is

If $\sin ^{ - 1}\frac{{2a}}{{1 + {a^2}}} - \cos ^{ - 1}\frac{{1 - {b^2}}}{{1 + {b^2}}} = \tan ^{ - 1}\frac{{2x}}{{1 - {x^2}}}$,then $x = $

The value of $\operatorname{cosec}^{-1}(\sqrt{2})+\cos ^{-1}\left(\frac{-1}{2}\right)-\sec ^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is equal to

The set of values of $x$ such that $\tan ^{-1}\left(\frac{x}{x-2}\right)-\tan ^{-1}\left(\frac{x}{2 x-1}\right)=\tan ^{-1}\left(\frac{2}{3}\right)$ is

The value of $\tan \left(\sin ^{-1}\left(\frac{3}{5}\right)+\tan ^{-1}\left(\frac{2}{3}\right)\right)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module