The sum of the first n terms of the series co | vedclass

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

Question

(D) The general term of the series is $T_r = \cot^{-1}(r^2 + r + 1)$.Using the identity $\cot^{-1} x = 	an^{-1} \left( \frac{1}{x} ight)$,we have $

Vedclass · Accepted Answer

All of these

Vedclass · Answer

(D) The general term of the series is $T_r = \cot^{-1}(r^2 + r + 1)$.Using the identity $\cot^{-1} x = 	an^{-1} \left( \frac{1}{x} ight)$,we have $T_r = 	an^{-1} \left( \frac{1}{r^2 + r + 1} ight)$.We can rewrite the denominator as $1 + r(r+1)$,so $T_r = 	an^{-1} \left( \frac{(r+1) - r}{1 + r(r+1)} ight)$.Using the formula $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$,we get $T_r = 	an^{-1}(r+1) - 	an^{-1}(r)$.The sum of the first $n$ terms is $S_n = \sum_{r=1}^{n} (	an^{-1}(r+1) - 	an^{-1}(r))$.This is a telescoping series: $S_n = (	an^{-1} 2 - 	an^{-1} 1) + (	an^{-1} 3 - 	an^{-1} 2) + \dots + (	an^{-1}(n+1) - 	an^{-1} n)$.$S_n = 	an^{-1}(n+1) - 	an^{-1} 1$.Using $	an^{-1} x - 	an^{-1} y = 	an^{-1} \left( \frac{x-y}{1+xy} ight)$,$S_n = 	an^{-1} \left( \frac{n+1-1}{1+(n+1)(1)} ight) = 	an^{-1} \left( \frac{n}{n+2} ight)$.Since $	an^{-1} x = \cot^{-1} \left( \frac{1}{x} ight)$,$S_n = \cot^{-1} \left( \frac{n+2}{n} ight)$.Thus,all options are correct.

The sum of the first $n$ terms of the series $\cot^{-1} 3 + \cot^{-1} 7 + \cot^{-1} 13 + \cot^{-1} 21 + \dots$ is given by:

Similar Questions

If $x \geq 1$,then $2 \tan^{-1} x + \sin^{-1} (\frac{2x}{1+x^2})$ is equal to:

If $2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x)$,then $x =$

$\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{8}$ has the value

Prove that $2 \tan ^{-1} \frac{1}{2} + \tan ^{-1} \frac{1}{7} = \tan ^{-1} \frac{31}{17}$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module