Question

If $n$ is the number of terms of the series ${\cot }^{-1}3,{\cot }^{-1}7,{\cot }^{-1}13,{\cot }^{-1}21\cdots ,$ whose sum is $\frac{1}{2}{\cos }^{-1}\left(\frac{24}{145}\right)$, then the value of $(n-5)$ is

Answer 1

$k^{th}$ term of the given sequence,

$T_k={\cot }^{-1}\left[\right(k+1{)}^2-k]={\tan }^{-1}(\frac{1}{k^2+k+1})={\tan }^{-1}(k+1)-{\tan }^{-1}k$
Sum of total $n$ terms $={\tan }^{-1}(n+1)-{\tan }^{-1}1={\tan }^{-1}(\frac{n}{n+2})$

$\Rightarrow \frac{1}{2}{\cos }^{-1}\left(\frac{24}{145}\right)=\frac{1}{2}{\tan }^{-1}\left(\frac{143}{24}\right)={\tan }^{-1}(\frac{n}{n+2})$

$\Rightarrow {\tan }^{-1}\left(\frac{143}{24}\right)=2{\tan }^{-1}(\frac{n}{n+2})={\tan }^{-1}(\frac{n(n+2)}{2(n+1)})$

$n=11\Rightarrow n-5=6$