Mathematics - Sequence & Series Question with Solution | TestHub

$\begin{array}{rl}& \text{ Given }{\tan }^{-1}\left(\frac{1}{1+1.2}\right)+{\tan }^{-1}\left(\frac{1}{1+2.3}\right)+\dots \\ & +{\tan }^{-1}\left(\frac{1}{1+n(n+1)}\right)={\tan }^{-1}(x)\end{array}$ $\begin{array}{rl}\Rightarrow {\tan }^{-1}\left(\frac{2-1}{1+1.2}\right)& +{\tan }^{-1}\left(\frac{3-2}{1+2.3}\right)+\dots \\ & +{\tan }^{-1}\left(\frac{n+1-n}{1+n(n+1)}\right)={\tan }^{-1}(x)\end{array}$ $\begin{array}{rl}& \Rightarrow {\tan }^{-1}(2)-{\tan }^{-1}(1)+{\tan }^{-1}(3)-{\tan }^{-1}(2)+\dots \\ & +{\tan }^{-1}(\mathrm{n}+1)-{\tan }^{-1}(\mathrm{n})={\tan }^{-1}(x)\end{array}$ $\Rightarrow {\tan }^{-1}(\mathrm{n}+1)-{\tan }^{-1}(1)={\tan }^{-1}(x)$ $\left\{\because {\tan }^{-1}(A)-{\tan }^{-1}(B)={\tan }^{-1}\left(\frac{A-B}{1+A\cdot B}\right)\right\}$ $\Rightarrow {\tan }^{-1}\left(\frac{n}{n+2}\right)={\tan }^{-1}(x)\Rightarrow x=\frac{n}{n+2}$