February 19th, 2018, 11:34 AM  #1 
Senior Member Joined: May 2014 From: Allentown PA USA Posts: 107 Thanks: 6 Math Focus: dynamical systen theory  Using Mathematical Induction
Dear My Math Forum Community: Prove: n^3  n is divisible by 3 for every positive integer n. [ Hint: use (k + 1)^3  (k + 1) = (k^3  k) + 3(k^2 + k). This proof is to illustrate the concept of mathematical induction. ] 
February 19th, 2018, 11:51 AM  #2 
Senior Member Joined: Sep 2016 From: USA Posts: 317 Thanks: 164 Math Focus: Dynamical systems, analytic function theory, numerics 
Base case is $n=1$ and 0 is clearly divisible by 3 so it holds. Now assume inductively that it holds for $k^3  k$. Applying the hint you have \[ (k + 1)^3  (k + 1) = (k^3  k) + 3(k^2 + k) \] so by your inductive assumption, the left summand is a multiple of 3. Clearly the right summand is as well and therefore, $(k+1)^3  (k+1)$ is a multiple of which and by induction, it holds for all $k \in \mathbb{N}$. 
February 19th, 2018, 11:51 AM  #3  
Senior Member Joined: Aug 2012 Posts: 1,780 Thanks: 482  Quote:
$n^3  n = n (n^2 1) = n (n + 1) (n  1)$ One of $n$, $n + 1$, and $n 1$ is divisible by $3$; therefore their product is.  
February 19th, 2018, 11:54 AM  #4 
Senior Member Joined: Sep 2016 From: USA Posts: 317 Thanks: 164 Math Focus: Dynamical systems, analytic function theory, numerics  I think the point is not to prove this "theorem" which is trivial. Rather, it is to learn how to apply induction, which the OP failed to do since he just asked someone to do it for him.

February 19th, 2018, 11:54 AM  #5 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,767 Thanks: 1017 Math Focus: Elementary mathematics and beyond 
$n^3n=n(n1)(n+1)=(n1)n(n+1).$ Now it's easily seen that this is divisible by 3. If you need to solve this by induction, you can start by telling us what your initial approach would be. 
February 20th, 2018, 09:45 AM  #6 
Math Team Joined: Jan 2015 From: Alabama Posts: 2,970 Thanks: 807 
If you are required to use induction, then you start, of course, by observing that when n= 1, this is 1 1= 0 which is divisible by 6. The next step is to show that "if, for some positive integer k, k^3 k is divisible by 6 then so is (k+1)^3 (k+1)". To do that, calculate (k+1)^3 (k+1)= k^3+ 3k^2+ 3k+ 1 k 1= k^3+ 3k^2+ 2k= (k^3 k)+ (3k^2+ 3k). Since k^3 k was divisible by 6 we can write this as 6n+ 3k(k+ 1) and we now can observe that, since k and k+ 1 are consecutive integers one of them is divisible by 2. 

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