JEE Main 2022 · 28 Jun · Shift 2 · Mathematics Q18

MathematicsSingle correct+4 / -1Question 18 of 90

The value of $\displaystyle\lim_{n\to\infty}6\tan\left\{\sum_{r=1}^n\tan^{-1}\left(\frac1{r^2+3r+3}\right)\right\}$ is equal to :

Answer and worked solution

Official answer: C

Telescoping. $\tan^{-1}\dfrac1{1+(r+1)(r+2)}=\tan^{-1}(r+2)-\tan^{-1}(r+1)$ , so the sum tends to $\frac\pi2-\tan^{-1}2$ .

Value. $\tan\left(\frac\pi2-\tan^{-1}2\right)=\frac12$ ; $6\cdot\frac12=3$ .

(C)

Topics:Inverse trigonometric functions Special series sums

JEE Main 2022 · 28 Jun · Shift 2, Mathematics question 18