Question

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

Answer 1

Answer: 11

$$\begin{gathered} T_1={\tan }^{-1}\frac{1}{3}={\tan }^{-1}2-{\tan }^{-1}1;T_2={\tan }^{-1}\frac{1}{7}={\tan }^{-1}3-{\tan }^{-1}2;T_3={\tan }^{-1}\frac{1}{13}={\tan }^{-1}4-{\tan }^{-1}3 \\ <br/>\text{ Clearly }{T}_n={\tan }^{-1}(n+1)-{\tan }^{-1}(n) \end{gathered}$$
$\text{ Hence }{S}_n={\tan }^{-1}(n+1)-{\tan }^{-1}1={\tan }^{-1}\left(\frac{n+1-1}{1+(n+1)\cdot 1}\right)=\left({\tan }^{-1}\frac{n}{n+2}\right)=\frac{1}{2}{\cos }^{-1}\left(\frac{24}{145}\right)$
$\Rightarrow 2\left({\tan }^{-1}\frac{n}{n+2}\right)={\cos }^{-1}\left(\frac{24}{145}\right)\left(\text{ Using }2{\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right)\forall x\ge 0\right)$
$\Rightarrow {\cos }^{-1}\left(\frac{2(n+1)}{n^2+2n+2}\right)={\cos }^{-1}\left(\frac{24}{145}\right)\Rightarrow \left(\frac{2(n+1)}{n^2+2n+2}\right)=\left(\frac{24}{145}\right)$
$\Rightarrow 12(n+1{)}^2-144(n+1)-(n+1)+12=0=((n+1)-12)(12(n+1)-1)=0$
$\therefore n+1=12,\frac{1}{12}\therefore n=11,\frac{-11}{12}\because n\in N\therefore n\ne \frac{-11}{12}$
Hence, $n=11$