JEE Main 2021 (Online) 26th August Evening ShiftQuestion 1
$\lim_{x\to2}\left(\sum_{n=1}^{9}\frac{x}{n(n+1)x^{2}+2(2n+1)x+4}\right)$ is equal to :
Limits & Continuity
Solution
$n(n+1)x^2+2(2n+1)x+4=(nx+2)((n+1)x+2)$. $\frac{x}{(nx+2)((n+1)x+2)}=\frac1{nx+2}-\frac1{(n+1)x+2}$, so the sum telescopes to $\frac1{x+2}-\frac1{10x+2}$. At $x=2$: $\frac14-\frac1{22}=\frac{9}{44}$. Option (C).