Mathematics - Sequence & Series Question with Solution | TestHub

Given, $y={\tan }^{-1}\frac{1}{x^2+x+1}$ $+{\tan }^{-1}\frac{1}{x^2+3x+3}+{\tan }^{-1}\frac{1}{x^2+5x+7}+\cdots \cdot$ to $n$ terms $={\tan }^{-1}\left\{\frac{1}{1+x(x+1)}\right\}+{\tan }^{-1}\left\{\frac{1}{1+(x+1)(x+2)}\right\}$ $+{\tan }^{-1}\left\{\frac{1}{1+(x+2)(x+3)}\right\}$ $+\dots \dots \dots +{\tan }^{-1}\left\{\frac{1}{1+(x+(n-1))(x+n)}\right\}$ $={\tan }^{-1}\left\{\frac{(x+1)-x}{1+(x+1)x}\right\}$ $+{\tan }^{-1}\left\{\frac{(x+2)-(x+1)}{1+(x+2)(x+1)}\right\}$ $+{\tan }^{-1}\left\{\frac{(x+3)-(x+2)}{1+(x+3)(x+2)}\right\}$ $+\dots .+{\tan }^{-1}\left(\frac{(x+n)-(x+n-1)}{1+(x+n)(x+n-1)}\right)$ $\therefore y=\left\{{\tan }^{-1}(x+1)-{\tan }^{-1}(x)\right\}$ $+\left\{{\tan }^{-1}(x+2)-{\tan }^{-1}(x+1)\right\}$ $+\left\{{\tan }^{-1}(x+3)-{\tan }^{-1}(x+2)\right\}$ $+\dots \dots ..+\left\{{\tan }^{-1}(x+n)-{\tan }^{-1}[x+(n-1)]\right\}$ So, $y={\tan }^{-1}(x+n)-{\tan }^{-1}(x)$ On differentiating both sides w.r.t. $x$, we get $\frac{dy}{dx}=\frac{1}{1+(x+n{)}^2}-\frac{1}{1+x^2}$