limx→2∑n=19xnn+1x2+22n+1x+4 is equal to

∑n=19xnn+1x2+22n+1x+4 =∑n=19xnx+2(n+1)x+2 =∑n=191nx+2-1(n+1)x+2 =1x+2-110x+2limx→2∑n=19xnn+1x2+22n+1x+4 =limx→21x+2−110x+2 =14-122=1888=944

limx→2∑n=19xnn+1x2+22n+1x+4 is equal to

∑n=19xnn+1x2+22n+1x+4 =∑n=19xnx+2(n+1)x+2 =∑n=191nx+2-1(n+1)x+2 =1x+2-110x+2limx→2∑n=19xnn+1x2+22n+1x+4 =limx→21x+2−110x+2 =14-122=1888=944

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$\lim_{x \to 2} (\sum_{n = 1}^{9} \frac{x}{n (n + 1) x^{2} + 2 (2 n + 1) x + 4})$ is equal to

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{5}$

$\frac{5}{24}$

$\frac{7}{36}$

$\frac{9}{44}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\sum_{n = 1}^{9} \frac{x}{n (n + 1) x^{2} + 2 (2 n + 1) x + 4} = \sum_{n = 1}^{9} \frac{x}{(n x + 2) ((n + 1) x + 2)} = \sum_{n = 1}^{9} (\frac{1}{n x + 2} - \frac{1}{(n + 1) x + 2}) = \frac{1}{x + 2} - \frac{1}{10 x + 2}$

$\lim_{x \to 2} (\sum_{n = 1}^{9} \frac{x}{n (n + 1) x^{2} + 2 (2 n + 1) x + 4}) = \lim_{x \to 2} \frac{1}{x + 2} - \frac{1}{10 x + 2} = \frac{1}{4} - \frac{1}{22} = \frac{18}{88} = \frac{9}{44}$