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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. |
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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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