Question

The value of $\cot \left[\sum _{n=1}^{23}{\cot }^{-1}\left(1+\sum _{k=1}^{n}2k\right)\right]$ is

• A. $\frac{23}{25}$
• B. $\frac{25}{23}$
• C. $\frac{23}{24}$
• D. $\frac{24}{23}$

Answer 1

The correct option is B

$\frac{25}{23}$

Explanation for the correct option

Step-1 Sum of AP series:

Given function is, $\cot \left[\sum _{n=1}^{23}{\cot }^{-1}\left(1+\sum _{k=1}^{n}2k\right)\right]$

Consider $\sum _{k=1}^{n}2k$
$\begin{array}{rcl}\sum _{k=1}^{n}2k& =& 2\left(1+2+\dots +n\right)\\ & =& 2\left[\frac{n\left(n+1\right)}{2}\right]\\ & =& n^2+n\end{array}$

Step-2 Sum of original series:

Thus,
$\cot \left[\sum _{n=1}^{23}{\cot }^{-1}\left(1+\sum _{k=1}^{n}2k\right)\right]=\cot \left[\sum _{n=1}^{23}{\cot }^{-1}\left(1+n^2+n\right)\right]$

Consider $\sum _{n=1}^{23}{\cot }^{-1}\left(1+n^2+n\right)$
We have,
$\begin{array}{cc}\sum _{n=1}^{23}{\cot }^{-1}\left(1+n^2+n\right)=\sum _{n=1}^{23}\left[{\tan }^{-1}\left(\frac{1}{1+n^2+n}\right)\right]& \left[\because co{t}^{-1}\left(x\right)=ta{n}^{-1}\left(\frac{1}{x}\right)\right]\\ =\sum _{n=1}^{23}\left[{\tan }^{-1}\left(\frac{1}{1+n\left(n+1\right)}\right)\right]& \\ =\sum _{n=1}^{23}\left[{\tan }^{-1}\left(\frac{\left(n+1\right)-n}{1+n\left(n+1\right)}\right)\right]& \\ =\sum _{n=1}^{23}\left[{\tan }^{-1}\left(n+1\right)-{\tan }^{-1}\left(n\right)\right]& \left[\because {\tan }^{-1}\left(a\right)-{\tan }^{-1}\left(b\right)={\tan }^{-1}\left(\frac{a-b}{1+\mathrm{ab}}\right)\right]\\ =\left[{\tan }^{-1}\left(2\right)-{\tan }^{-1}\left(1\right)+{\tan }^{-1}\left(3\right)-{\tan }^{-1}\left(2\right)+\dots +{\tan }^{-1}\left(24\right)-{\tan }^{-1}\left(23\right)\right]& \\ =\left[{\tan }^{-1}\left(24\right)-{\tan }^{-1}\left(1\right)\right]& \\ ={\tan }^{-1}\left(\frac{24-1}{1+24\times 1}\right)& \\ ={\tan }^{-1}\left(\frac{23}{25}\right)& \end{array}$

Thus,
$\begin{array}{rcl}& & \begin{array}{cc}\cot \left[\sum _{\mathrm{n}=1}^{23}{\cot }^{-1}\left(1+{\mathrm{n}}^2+\mathrm{n}\right)\right]=\cot \left[{\tan }^{-1}\left(\frac{23}{25}\right)\right]& \\ =\cot \left[{\cot }^{-1}\left(\frac{25}{23}\right)\right]& \left[\because {\tan }^{-1}\left(\mathrm{x}\right)={\cot }^{-1}\left(\frac{1}{\mathrm{x}}\right)\right]\\ =\frac{25}{23}& \left[\because \cot \left({\cot }^{-1}\left(\mathrm{x}\right)\right)=\mathrm{x}\right]\end{array}\end{array}$

Therefore, $\cot \left[\sum _{n=1}^{23}{\cot }^{-1}\left(1+\sum _{k=1}^{n}2k\right)\right]=\frac{25}{23}$

Hence, option(B) is correct.