Question

If $\tan \left(\sum _{r=1}^n{\tan }^{-1}\left(\frac{2r-1}{\left(r^2+r+1\right)\left(r^2-r+1\right)-2r^3}\right)\right)=961$, then the value of $n$ is equal to

• A. 31
• B. 30
• C. 30.
• D. 61

Answer 1

Answer: (A)

$\sum _{r=1}^n{\tan }^{-1}\left(\frac{2r-1}{\left(r^2+r+1\right)\left(r^2-r+1\right)-2r^3}\right)=\sum _{r=1}^n{\tan }^{-1}\left(\frac{2r-1}{r^4+r^2+1-2r^3}\right)$
$=\sum _{r=1}^n{\tan }^{-1}\left(\frac{2r-1}{1+r^2\left(r^2-2r+1\right)}\right)=\sum _{r=1}^n{\tan }^{-1}\left(\frac{r^2-(r-1{)}^2}{1+r^2(r-1{)}^2}\right)=\sum _{r=1}^n\left({\tan }^{-1}{r}^2-{\tan }^{-1}(r-1{)}^2\right)$
$={\tan }^{-1}\left(n^2\right)$
$\text{ Now }\tan \left({\tan }^{-1}\left(n^2\right)\right)=961\Rightarrow n^2=961$
$\Rightarrow n=31$