Question

If $\displaystyle \sum _{n = 1}^{\infty }\left(tan\right)^{- 1}\left(\frac{1}{7 - 5 n + n^{2}}\right)=\frac{\pi }{2}+\left(tan\right)^{- 1}c$ , then $c$ is equal to

• A. $2$
• B. $3$
• C. $17$
• D. $\frac{5}{2}$

Answer 1

Answer: (A)

$\sum_{n=1}^{\infty} \tan ^{-1}\left(\frac{1}{7-5 n+n^{2}}\right)=\sum_{n=1}^{\infty} \tan ^{-1} \frac{(n-2)-(n-3)}{1+(n-2)(n-3)}$
$=\sum_{ n =1}^{\infty}\left[\tan ^{-1}( n -2)-\tan ^{-1}( n -3)\right]$
$=\frac{\pi}{2}-\tan ^{-1}(-2)=\frac{\pi}{2}+\tan ^{-1} 2=\frac{ a \pi}{ b }+\tan ^{-1} c$
$a =1, \quad b =2, \quad c =2$