Answer to Question #101081 in Discrete Mathematics for ludnor

Question

Use mathematical induction to prove that ∀n∈N

P(n):1·2·3+2·3·4+···+n(n+1)(n+2)=n(n+1)(n+2)(n+3)/4

Answer 1

We use induction method to prove that p(n) holds for all positive integers n, where

p(n): 1*2*3+....+n(n+1)(n+2) = n(n+1)(n+2)(n+3)/4

Induction base case

p(1): 1*2*3 = 1(1+1)(1+2)(1+3)/4 which is equivalent with 1*2*3=1*2*3*4/4 or 1*2*3=1*2*3

Thus base case holds.

Induction general step

We assume that p(k) holds for n>1 and we must prove that p(k) also holds for n+1.

p(n+1) claims:

1*2*3+....+(n+1)(n+2)(n+3)=(n+1)(n+2)(n+3)(n+4)/4

But 1*2*3+....+(n+1)(n+2)(n+3) =[1*2*3+....+n(n+1)(n+2)]+(n+1)(n+2)(n+3)

For the bracket we use the inductive hypothesis which holds for k=n and plug the expression

to obtain:

1*2*3+....+(n+1)(n+2)(n+3)= n(n+1)(n+2)(n+3)/4 + (n+1)(n+2)(n+3)=

=(n+1)(n+2)(n+3)(n/4 +1) =(n+1)(n+2)(n+3)(n+4)/4 which proves the induction claim.

In conclusion, the induction proof is now complete and the identity holds for all positive integers n, as required.