Answer:
Option B
Explanation:
Concept involved difference series , when ever , we have summation involving more than 3 terms we should always covert into diferences
e.g, $\sum_{r=1}^{5}\frac{1}{r(r+1)}=\frac{1}{1.(2)}+\frac{1}{2.(3)}+\frac{1}{3.(4)}+\frac{1}{4.5}+\frac{1}{5.6}$
$=\frac{2-1}{1.(2)}+\frac{3-2}{2.(3)}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+\frac{6-5}{5.6}$
$=(1-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{6})=1-\frac{1}{6}=\frac{5}{6}$
Situation Analysis
Convert into
$\tan^{-1} x-\tan^{-1} y=\tan^{-1} \left(\frac{x-y}{1+xy}\right)$
$\cot\left(\sum_{n=1}^{23}\cot^{-1}\left(1+\sum_{n=1}^{n}2k\right)\right)$
$\cot\left(\sum_{n=1}^{23}\cot^{-1}(1+2+4+6+8...+2n)\right)$
$\Rightarrow\cot\left(\sum_{n=1}^{23}\cot^{-1}(1+n(n+1))\right)$
$\Rightarrow\cot\left(\sum_{n=1}^{23}\tan^{-1}\frac{1}{1+n(n+1)}\right)$
$\Rightarrow\cot\left(\sum_{n=1}^{23}\tan^{-1}\frac{(n+1)-n}{1+n(n+1)}\right)$
$\Rightarrow\cot\left(\sum_{n=1}^{23}\tan^{-1}(n+)-\tan^{-1}ln n\right)$
$\Rightarrow\cot\left(\tan^{-1}2-\tan^{-1}1\right)+\left(\tan^{-1}3-\tan^{-1}2\right)+\left(\tan^{-1}4-\tan^{-1}3\right)$+ $.............. +\left(\tan^{-1}24-\tan^{-1}23\right)$
$\Rightarrow\cot\left(\tan^{-1}24-\tan^{-1}1\right)$
$\Rightarrow\cot\left(\tan^{-1}\frac{24-1}{1+24-(1)}\right)$
$\Rightarrow\cot\left(\tan^{-1}\frac{23}{25}\right)$
= $\cot\left(\cot^{-1}\frac{25}{23}\right)$
=$\frac{25}{23}$
