Question

How many terms of the sequence $\cot ^{-1} 3, \cot ^{-1} 7, \cot ^{-1} 13, \cot ^{-1} 21, \ldots \ldots \ldots \ldots \ldots .$. must be taken to have their sum equal to $\frac{1}{2} \cos ^{-1}\left(\frac{24}{145}\right)$.

Answer 1

$T _1=\tan ^{-1} \frac{1}{3}=\tan ^{-1} 2-\tan ^{-1} 1 ; T _2=\tan ^{-1} \frac{1}{7}=\tan ^{-1} 3-\tan ^{-1} 2 ; T _3=\tan ^{-1} \frac{1}{13}=\tan ^{-1} 4-\tan ^{-1} 3 \\
\text { Clearly } T _{ n }=\tan ^{-1}( n +1)-\tan ^{-1}( n ) $
$\text { Hence } S _{ n }=\tan ^{-1}( n +1)-\tan ^{-1} 1=\tan ^{-1}\left(\frac{ n +1-1}{1+( n +1) \cdot 1}\right)=\left(\tan ^{-1} \frac{ n }{ n +2}\right)=\frac{1}{2} \cos ^{-1}\left(\frac{24}{145}\right) $
$\Rightarrow 2\left(\tan ^{-1} \frac{ n }{ n +2}\right)=\cos ^{-1}\left(\frac{24}{145}\right) \left(\text { Using } 2 \tan ^{-1} x =\cos ^{-1}\left(\frac{1- x ^2}{1+ x ^2}\right) \forall x \geq 0\right) $
$\Rightarrow \cos ^{-1}\left(\frac{2( n +1)}{ n ^2+2 n +2}\right)=\cos ^{-1}\left(\frac{24}{145}\right) \Rightarrow\left(\frac{2( n +1)}{ n ^2+2 n +2}\right)=\left(\frac{24}{145}\right)$
$\Rightarrow 12( n +1)^2-144( n +1)-( n +1)+12=0=(( n +1)-12)(12( n +1)-1)=0 $
$\therefore n +1=12, \frac{1}{12} \therefore n =11, \frac{-11}{12} \because n \in N \therefore n \neq \frac{-11}{12}$
Hence, $n=11$