1+n2+n(n−1)2⋅4+n(n−1)(n−2)2⋅4⋅6+………+∞=

1+n2+n(n−1)2.4+…=1+n12+nn-12!122+.....=1+12n=32n1+n3+n(n+1)3.6+…=1+n113+nn+12!132+......=1-13-n=32n

1+n2+n(n−1)2⋅4+n(n−1)(n−2)2⋅4⋅6+………+∞=

1+n2+n(n−1)2.4+…=1+n12+nn-12!122+.....=1+12n=32n1+n3+n(n+1)3.6+…=1+n113+nn+12!132+......=1-13-n=32n

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

$1 + \frac{n}{2} + \frac{n (n - 1)}{2 \cdot 4} + \frac{n (n - 1) (n - 2)}{2 \cdot 4 \cdot 6} + \dots \dots \dots + \infty =$

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

$1 + \frac{n}{3} + \frac{n (n + 1)}{3 .6} + \frac{n (n + 1) (n + 2)}{3 .6.9} + . . .$

$1 + \frac{n}{3} + \frac{n (n + 2)}{3 .6} + \frac{n (n + 1) (n + 2)}{3 .6.9} + \dots$

$1 + \frac{n}{3} + \frac{n (n + 2)}{3 .6} + \frac{n (n + 1) (n + 1)}{3 .6.9} + \dots$

$1 + \frac{n}{3} + \frac{n (n - 1)}{3 .6} + \frac{n (n + 1) (n + 1)}{3 .6.9} + \dots$

answer is C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\begin{array}{l} 1 + \frac{n}{2} + \frac{n (n - 1)}{2 .4} + \dots = 1 + n (\frac{1}{2}) + \frac{n (n - 1)}{2!} {(\frac{1}{2})}^{2} + . . . . . = {(1 + \frac{1}{2})}^{n} = {(\frac{3}{2})}^{n} \\ 1 + \frac{n}{3} + \frac{n (n + 1)}{3 .6} + \dots = 1 + \frac{n}{1} (\frac{1}{3}) + \frac{n (n + 1)}{2!} {(\frac{1}{3})}^{2} + . . . . . . = {(1 - \frac{1}{3})}^{- n} = {(\frac{3}{2})}^{n} \end{array}$