Question

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

• A. 31
• B. 30
• C. 30.
• D. 61

Answer 1

Answer: (A)

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