JEE Main 2022 · 27 Jun · Shift 2 · Mathematics Q18

MathematicsSingle correct+4 / -1Question 18 of 90

The value of $\cot\left(\displaystyle\sum_{n=1}^{50}\tan^{-1}\left(\frac1{1+n+n^2}\right)\right)$ is

\dfrac{26}{25}

\dfrac{25}{26}

\dfrac{50}{51}

\dfrac{52}{51}

Answer and worked solution

Official answer: A

Telescoping. $\tan^{-1}\dfrac1{1+n+n^2}=\tan^{-1}(n+1)-\tan^{-1}n$ , so the sum is $\tan^{-1}51-\tan^{-1}1=\tan^{-1}\dfrac{50}{52}$ .

Cot. $\dfrac{52}{50}=\dfrac{26}{25}$ .

(A)

Topics:Inverse trigonometric functions Special series sums

JEE Main 2022 · 27 Jun · Shift 2, Mathematics question 18