Question

Using principle of mathematical induction.Prove that n(n+1)(n+2) is a multiple of 6

Answer 1

Answer:

Trivial case: n=0n=0. Observe that 40–140–1 is divisible by 33. Done.

Inductive case: Assume 4n−14n−1 is divisible by 33. Show that this implies 4n+1−14n+1−1 is divisible by 33.

4n+1−1=4⋅4n−14n+1−1=4⋅4n−1

=(3+1)4n−1=(3+1)4n−1

=3⋅4n+4n−1=3⋅4n+4n−1

=(3⋅4n)+(4n−1)=(3⋅4n)+(4n−1)

The first term is divisible by 33, since it has three as a factor.

The second term is divisible by 33 by the inductive hypothesis.

So the whole thing is divisible by 33.

By induction,

4n−14n−1 is divisible by 33 for any nonnegative integer nn.