Question

If $\sin \left(\sum _{r=1}^{40}{\cot }^{-1}\right(1+r+r^2\left)\right)=\frac{a}{b},$ where $a$ and $b$ are co-prime, then $a+b$

• A. is a prime number
• B. has odd number of factors
• C. is divisible by $5$
• D. None of the above

Answer 1

The correct option is B has odd number of factors

$S=\sum _{r=1}^{40}{\cot }^{-1}(1+r+r^2)=\sum _{r=1}^{40}{\tan }^{-1}\left(\frac{1}{1+r+r^2}\right)=\sum _{r=1}^{40}{\tan }^{-1}\left(\frac{(r+1)-\left(r\right)}{1+r+r^2}\right)=\sum _{r=1}^{40}\left({\tan }^{-1}\right(r+1)-{\tan }^{-1}(r\left)\right)={\tan }^{-1}\left(41\right)-{\tan }^{-1}\left(1\right)={\tan }^{-1}\frac{20}{21}={\sin }^{-1}\frac{20}{29}\therefore \frac{a}{b}=\frac{20}{29}$

$a+b=49$