The value of n ( k=1n k3 + 6k2 + 11k + 5(k+3)! ) is :
(1) 5/3
k3 + 6k2 + 11k + 5 = (k + 1)(k + 2)(k + 3) - 1
k3 + 6k2 + 11k + 5(k + 3)! = 1k! - 1(k + 3)!
Limit = k=1 1k! - k=1 1(k + 3)!
= (e - 1) - (e - 1 - 1 - 12 - 16 )
= 11! + 12! + 13! = 53
Answer: (1)
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