Show n^3 + 11n Divisible by 6Date: 12/11/2002 at 00:41:56 From: Jasmine Subject: Algebra If n is a natural number, show that for all values of n, (n^3+11n) is divisible by 6. Date: 12/11/2002 at 10:33:52 From: Doctor Ian Subject: Re: Algebra Hi Jasmine, I would probably try to use induction. That is, I know that the formula works for small values of n: n=1, n=2, and so on. So what I want to show is that if it works for any particular n, it must also work for the next value of n, i.e., (n+1). That is, If n^3 + 11n is divisible by 6, then (n+1)^3 + 11(n+1) is also divisible by 6. Now, (n+1)^3 + 11(n+1) = (n^3 + 3n^2 + 3n + 1) + (11n + 11) = (n^3 + 11n) + (3n^2 + 3n + 12) We're already assuming that the first part, (n^3 + 11n) is divisible by 6; so now we have a smaller problem, which is to show that the second part, 3n^2 + 3n + 12 is always divisible by 6. (Hint: Show that it's divisible by both 2 and 3.) Can you take it from here? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/ Date: 12/11/2002 at 19:47:05 From: Jasmine Subject: Thank you (Algebra) Got the hint. Thank you so much. You have been a great help. Jasmine |
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