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Q: Discrete Mathematics ( Answered,   0 Comments )
Question  
Subject: Discrete Mathematics
Category: Computers
Asked by: math01-ga
List Price: $2.00
Posted: 17 Nov 2002 10:32 PST
Expires: 17 Dec 2002 10:32 PST
Question ID: 109391
7. Use mathematical induction to show that 3 divides n^3 + 2n whenever
n is a nonnegative integer.
Answer  
Subject: Re: Discrete Mathematics
Answered By: livioflores-ga on 17 Nov 2002 21:26 PST
 
Hi math01!!

For n=1
n^3 + 2n = 1^3 + 2x1 = 1 + 2 = 3  th proposition is true.

Now we will assume that the proposition is true for some nonnegative
integer n, then:

n^3 + 2n = 3k, where k is a nonnegative integer.

Then we must prove the proposition for the number (n+1):

(n+1)^3 + 2(n+1) = (n^3 + 3n^2 + 3n + 1)+ 2n + 2 =
                 = n^3 + 3n^2 + 3n + 2n + 3 =
                 = (n^3 + 2n)+ 3(n^2 + 3n + 3)=  (If we do (n^2 + 3n +
3)= m)
                 = 3k + 3m = 3(k+m)  where (k+m) is a nonnegative
integer.

Now the proposition is demonstrated.

I use my own knowledge to do this, if you need some clarification
please post a request for it.

Regards
livioflores-ga
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