_ n , _ r = 0 ^n ^ - 1 ( 1 + r + r^2 ) n is equal to - Vidyadip

MCQ

$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum\limits_{r = 0}^n {{{\tan }^{ - 1}}\left( {1 + r + {r^2}} \right)} }}{n}$ is equal to

A
$1$
B
$2$
C
$\frac {\pi}{4}$
✓
$\frac {\pi}{2}$

✓

Answer

Correct option: D.

$\frac {\pi}{2}$

d
$\sum\limits_{r = 0}^n {{{\tan }^{ - 1}}} \left( {1 + r + {r^2}} \right)$

$ = \sum\limits_{r = 0}^n {\frac{\pi }{2} - {{\tan }^{ - 1}}} \frac{1}{{1 + r + {r^2}}}$

$ = \frac{{n\pi }}{2} - \sum\limits_{r = 0}^n {{{\tan }^{ - 1}}} \frac{{\left( {r + 1} \right) - r}}{{1 + r\left( {r + 1} \right)}}$

$ = \frac{{n\pi }}{2} - \sum\limits_{r = 0}^n {\left( {{{\tan }^{ - 1}}\left( {r + 1} \right) - {{\tan }^{ - 1}}r} \right)} $

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