Prove that n^3 + 2n is Divisible by 3Date: 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/ |
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