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Q: MATH INDUCTION ( Answered 5 out of 5 stars,   3 Comments )
Question  
Subject: MATH INDUCTION
Category: Science > Math
Asked by: judgementday-ga
List Price: $2.00
Posted: 22 Jul 2002 01:49 PDT
Expires: 21 Aug 2002 01:49 PDT
Question ID: 43645
Prove by induction that the product of every three consecutive even
integers is divisible by 48. ex: 2 * 4 * 6 is divisble by 48

Request for Question Clarification by calebu2-ga on 25 Jul 2002 09:03 PDT
judgementday - can i repost my comment as an answer to your question?
It proves what you needed it to.

Regards

calebu2-ga
Answer  
Subject: Re: MATH INDUCTION
Answered By: calebu2-ga on 26 Jul 2002 11:35 PDT
Rated:5 out of 5 stars
 
Mr. Day,

Before I do a repeat of my answer - here are a few cool links about
induction (just to keep your interest flowing) :

Induction Principle from Mathworld@Wolfram
http://mathworld.wolfram.com/InductionPrinciple.html

Course notes at USC on Mathematical Induction
http://www.math.sc.edu/~sumner/numbertheory/induction/Induction.html

Search term used on google.com : "proof by induction"

--------------

Prove : For every n, the product of the numbers 2n, 2n + 2 and 2n + 4
is divisible by 48.
 
ie. 48 divides 2n(2n+2)(2n+4) = 8n(n+1)(n+2). 
 
Here's the induction part: 
 
Let n = 1 
 
2n(2n+2)(2n+4) = 2*4*6 = 48. True for n = 1. 
 
Assume it is true for n = k. ie. 48 divides 8k(k+1)(k+2). ie. 6
divides k(k+1)(k+2).
 
Consider n = k+1 : 
 
We need 48 to divide 8(k+1)(k+2)(k+3). 
 
Ie. 6 to divide (k+1)(k+2)(k+3). 
 
Now (k+1)(k+2)(k+3) = k(k+1)(k+2) + 3(k+1)(k+2) 
 
By induction 6 divides k(k+1)(k+2) 
and if 2 does not divide k+1, it must divide k+2 (either k+1 is even
or k+2 is even). Therefore 6 = 3 * 2 must divide 3(k+1)(k+2).
 
Therefore 48 divides 8(k+1)(k+2)(k+3). 
 
Hence, by induction, for every .n, the product of the numbers 2n, 2n +
2 and 2n + 4 is divisible by 48

Good luck with the rest of your proofs!

calebu2-ga
judgementday-ga rated this answer:5 out of 5 stars
great job. Used correct mathematical induction to solve problem

Comments  
Subject: Re: MATH INDUCTION
From: calebu2-ga on 22 Jul 2002 06:45 PDT
 
I think the induction proof goes something like this :

Prove : For every n, the product of the numbers 2n, 2n + 2 and 2n + 4
is divisible by 48.

ie. 48 divides 2n(2n+2)(2n+4) = 8n(n+1)(n+2).

Here's the induction part:

Let n = 1

2n(2n+2)(2n+4) = 2*4*6 = 48. True for n = 1.

Assume it is true for n = k. ie. 48 divides 8k(k+1)(k+2). ie. 6
divides k(k+1)(k+2).

Consider n = k+1 :

We need 48 to divide 8(k+1)(k+2)(k+3).

Ie. 6 to divide (k+1)(k+2)(k+3).

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

By induction 6 divides k(k+1)(k+2)
and if 2 does not divide k+1, it must divide k+2 (either k+1 is even
or k+2 is even). Therefore 6 = 3 * 2 must divide 3(k+1)(k+2).

Therefore 48 divides 8(k+1)(k+2)(k+3).

Hence, by induction, for every n, the product of the numbers 2n, 2n +
2 and 2n + 4 is divisible by 48.

I'm sure that could be written more clearly, but that's how an
induction proof goes. In some ways it is obvious that any 3
consecutive even numbers must contain the factors 2, 2, 2, 2 and 3 -
but just observing that does not constitute a proof - and it an isn't
inductive proof. Other than that, a good concise answer.

calebu2-ga
Subject: Re: MATH INDUCTION
From: cheef-ga on 22 Jul 2002 07:08 PDT
 
The "Answer" (gosh, I hate even saying those two words with all the
Allen Iverson junk in the news) fails to use induction.  calebu2's
"Comment" does the job using induction.
Subject: Re: MATH INDUCTION
From: judgementday-ga on 25 Jul 2002 23:04 PDT
 
yes please do. This is a first time posting a question so I did not
know how to contact calebu to give him the 2 bucks (a well earned two
bucks).

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