Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

The general term of the series is $T_n={\tan }^{-1}\left(\frac{2n}{n^4+n^2+1}\right)$. We can factor the denominator as $n^4+n^2+1=(n^2+1{)}^2-n^2=(n^2+1-n)(n^2+1+n)$. The numerator $2n$ can be written as $(n^2+n+1)-(n^2-n+1)$. Thus, $T_n={\tan }^{-1}\left(\frac{(n^2+n+1)-(n^2-n+1)}{1+(n^2+n+1)(n^2-n+1)}\right)$. Using the identity ${\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)$, we can write $T_n={\tan }^{-1}(n^2+n+1)-{\tan }^{-1}(n^2-n+1)$. Let $f(n)={\tan }^{-1}(n^2-n+1)$. Then $f(n+1)={\tan }^{-1}((n+1{)}^2-(n+1)+1)={\tan }^{-1}(n^2+2n+1-n-1+1)={\tan }^{-1}(n^2+n+1)$. So, $T_n=f(n+1)-f(n)$. This is a telescoping series. The sum $S_{10}={\sum }_{n=1}^{10}[f(n+1)-f(n)]$ is:$S_{10}=[f(2)-f(1)]+[f(3)-f(2)]+\cdots +[f(11)-f(10)]=f(11)-f(1)$. Now, we calculate $f(1)$ and $f(11)$:$f(1)={\tan }^{-1}(1^2-1+1)={\tan }^{-1}(1)=\frac{\pi }{4}$.$f(11)={\tan }^{-1}(11^2-11+1)={\tan }^{-1}(121-11+1)={\tan }^{-1}(111)$. Therefore, the sum $S_{10}={\tan }^{-1}(111)-\frac{\pi }{4}$. The final answer is $\boxed{{\tan }^{-1}(111)-\frac{\pi }{4}}$.