IIT JEE , AIEEE Forum - Mathematics , Algebra

sneha_91

$\sum_{a}^{b}$ r(r+1)(r+2) where a = r=1 and b=infinity

a)4(ⁿ⁺³c₂) b)6(ⁿ⁺³c₄) find the answer...explain the solution

allamraju

sneha_91

u r right it is till n

sneha_91

please tell me how to begin this problem

allamraju

iitkgp_bipin

Subjective approach has already been discussed by allamraju.

Objective approach : put n=1

summation = 1(1+1)(1+2) = 6

for n=1 : option (a) = 4(⁴C₂) = 24

for n=1 : option (b) = 6(⁴C₄) = 6

since option (b) matches with original summation, correct answer is option (b).

hsbhatt

Another usually used method is to write:

$\sum_{r=1}^n r(r+1)(r+2) = \frac{1}{4} \sum_{r=1}^n [(r+3)(r+2)(r+1)r - (r+2)(r+1)r(r-1)]$

which is a telescopic sum which easily computes to

$\frac {(n+3)(n+2)(n+1)n}{4} = 6 \ \frac {(n+3)(n+2)(n+1)n}{1.2.3.4} = 6 \binom{n+3}{4}$