Q. Sum of first $n$ terms in the series $\cot^{- 1}{3 + \cot^{- 1}{7 + \cot^{- 1}{13 + \cot^{- 1}}{21 +}\ldots}}$ is given by
A) $\tan^{- 1}\left( \frac{n}{n + 2} \right)$
B) $\cot^{- 1}\left( \frac{n + 2}{n} \right)$
C) $\tan^{- 1}{(n + 1) - \tan^{- 1}1}$
D) All of these
Explanation
(b)
Let $S = 3 + 7 + 13 + 21 + ... + T_{n}$
$\Rightarrow T_{n} = n^{2} + n + 1$
Let $T_{r} = \cot^{- 1}{(r^{2} + r + 1)}$
$= \tan^{- 1}{(r + 1) - \tan^{- 1}r}$
Put $r = 1,\ 2,\ldots,n$
$T_{1} = \tan^{- 1}{2 - \tan^{- 1}1}$
$T_{2} = \tan^{- 1}{3 - \tan^{- 1}2}$
$\cdots\ \cdots\ \cdots$
$\cdots\ \cdots\ \cdots$
$T_{n} = \tan^{- 1}{(n + 1) - \tan^{- 1}n}$
On adding all these, we get
$T_{1} + T_{2} + ... + T_{n} = \tan^{- 1}{(n + 1) - \tan^{- 1}{1 = \tan^{- 1}{\left( \frac{n}{n + 2} \right) = \cot^{- 1}\left( \frac{n + 2}{n} \right)}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.