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January 31st, 2018, 11:29 PM  #1 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,766 Thanks: 121 Math Focus: Trigonometry and Logarithm  [ASK] Prove Using Induction
Prove that n(n + 1)(n + 2) is divisible by 6 for any integer n What I have done so far: For n = 1 1(1 + 1)(1 + 2) = 1(2)(3) = 6(1) = 6 6 is divisible by 6 (TRUE) For n = k k(k + 1)(k + 2) is divisible 6 (ASSUMED AS TRUE) For n = k + 1 (k + 1)(k + 2)(k + 3) = k(k + 1)(k + 2) + 3(k + 1)(k + 2) What am I supposed to do from here onward? Yes, I know that: k(k + 1)(k + 2) is divisible by 6 (k + 1) and (k + 2) are subsequent numbers so that its product is even or divisible by 2. Because (k + 1)(k + 2) is divisible by 2 then 3(k + 1)(k + 2) is divisible by 6. Because k(k + 1)(k + 2) is divisible by 6 and 3(k + 1)(k + 2) is divisible by 6 then k(k + 1)(k + 2) + 3(k + 1)(k + 2) is divisible by 6 (PROVEN TRUE) However, I am not sure if that follows the standard procedure of mathematical induction. Can someone please show me the "proper" "formal" way using mathematical induction continued from the bolded part? 
January 31st, 2018, 11:32 PM  #2  
Senior Member Joined: Aug 2012 Posts: 1,702 Thanks: 449  Quote:
And it's divisible by 2 since at least one of the numbers must be even. So it's divisible by 6. This is much easier than induction.  
January 31st, 2018, 11:37 PM  #3 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,766 Thanks: 121 Math Focus: Trigonometry and Logarithm 
Unfortunately, we're making math textbook and this problem is in the mathematical induction chapter.

February 1st, 2018, 02:41 AM  #4 
Senior Member Joined: Oct 2009 Posts: 229 Thanks: 81 
Yeah, but if you're going to put a problem in an induction chapter that is clearly and obviously easiest to solve without doing induction, the book is just bad. I can just imagine the book being used in the classroom. The teacher assigns the problem. The clever student does it the obvious way and doesn't use induction and loses points because of it... 
February 1st, 2018, 04:47 PM  #5 
Senior Member Joined: Nov 2010 From: Indonesia Posts: 1,766 Thanks: 121 Math Focus: Trigonometry and Logarithm 
Ugh, that was... harsh. But I admit you do have a point.

February 1st, 2018, 05:04 PM  #6 
Senior Member Joined: Sep 2016 From: USA Posts: 276 Thanks: 141 Math Focus: Dynamical systems, analytic function theory, numerics 
I disagree with the previous comments. The fact that this "theorem" is obvious by a direct proof has nothing to do with it. The purpose of the exercise has nothing to do with whether or not the product is divisible by 6 and everything to do with practicing the method of proof by induction on particularly easy examples. To answer your question, your reasoning beneath the bold statement is perfectly fine. Beginning at the bolded statement this is one way to finish: By a direct expansion we have \[(k + 1)(k + 2)(k + 3) = k(k + 1)(k + 2) + 3(k + 1)(k + 2). \] By our inductive hypothesis, there exists an integer, $m$ such that \[k(k + 1)(k + 2) = 6m.\] Now, $k+1,k+2$ are consecutive integers and therefore one of these is even implying that there exists an integer $n$ such that $(k+1)(k+2) = 2n$. It follows that $3(k + 1)(k + 2) = 6n$. Taken together, we have \[(k + 1)(k + 2)(k + 3) = 6(n+m)\] and we conclude from the principle of mathematical induction that $N(N+1)(N+2)$ is a multiple of 6 for every natural number $N$. 
February 1st, 2018, 05:07 PM  #7 
Senior Member Joined: Aug 2012 Posts: 1,702 Thanks: 449  It's harder than it looks to find problems that truly require induction. A lot of problems students ask about online are like this. Induction problems that are easier done without induction. No wonder the students are confused. I propose a rule of thumb. If an undergrad math major could look at the problem and see the answer in a few seconds ... that's not a good candidate for an induction problem. Last edited by Maschke; February 1st, 2018 at 05:12 PM. 

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