Question

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

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

Answer 1

Answer: (A)

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