Prove that n^3 + 2n is Divisible by 3

Date: 04/15/2001 at 20:34:11
From: Jenny
Subject: Mathematical Induction

Prove by mathematical induction that n^3 + 2n is divisible by 3 for 
all positive integers.

Thanks.

Date: 04/15/2001 at 22:55:46
From: Doctor Jaffee
Subject: Re: Mathematical Induction

Hi Jenny,

Proving a mathematical statement by mathematical induction is much 
like proving that you can climb as high as you want on a ladder of 
infinite height. 

You have to be able to reach the first step of the ladder. If you 
can't get to the first step, it doesn't make any difference how good 
your climbing skills are; you can't go anywhere.

Next, can you show that no matter what step of the ladder you are on, 
you can always get to the next higher one?

If you can do both of these things, you have successfully proved that 
you can climb as high as you want.

Proving a mathematical statement by mathematical induction is 
similar. First, can you get to the first step; that is, can you prove 
that n^3 + 2n is divisible by 3 for n = 1? That should be pretty easy. 
Just substitute 1 for n and check it out.

Next, if you are on any step of the ladder, can you prove that you can 
get to the next one? This is just like saying "If we assume that the 
statement is true for the case when n = k, can we prove that it is 
true for n = k+1?

In other words, if we assume k^3 + 2k is divisible by 3, can we prove 
that (k+1)^3 + 2(k+1) is divisible by 3? Keep in mind that:

     (k+1)^3 + 2(k+1) = (k^3 + 3k^2 + 3k + 1) + 2k + 2

Rearrange terms and we have (k^3 + 2k) + 3k^2 + 3k + 3.

Can you take it from there? You should be able to prove that the 
expression above is divisible by 3.

Give it a try and if you want to check your answer with me, write 
back. If you are having difficulties, let me know and I'll try to help 
you some more.

Good luck.

- Doctor Jaffee, The Math Forum
  http://mathforum.org/dr.math/