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anonymous

  • 5 years ago

I am stuck with a proof by induction on doing the factoring can someone help?

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  1. anonymous
    • 5 years ago
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    I need to prove that \[(n^3-n)(n+2) \] is divisible by 12. I have gotten to the point of replacing n with n+1 but I cant get a factor of 12 anywhere

  2. anonymous
    • 5 years ago
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    I have an answer...bear with me, it may take a little time to write out.

  3. anonymous
    • 5 years ago
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    When you assume the first statement as true, S(k), write it in a more amenable form:\[(k^3-k)(k+2)=k(k^2-1)(k+2)=k(k+1)(k-1)(k+2)\] and we assume that this equals \[12j\] for some j an integer. Now we want to show S(k+1) is true given our assumption. Then \[(k+1)(k+2)(k)(k+3)\] substituting k+1 into the first equation. From S(k), we know \[k(k+1)(k+2)=12j/(k-1)\] and so statement S(k+1) becomes, \[12j(k+3)/(k-1)\]Since 12 is a factor, S(k+1) is divisible by 12. Of course, you have to clean it up and go through the motions in a little more detail (e.g. testing the first case). Please 'fan' me if this helps :)

  4. anonymous
    • 5 years ago
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    PS the first equation is missing the end of the (k+2) factor.

  5. anonymous
    • 5 years ago
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    i am not sure why you chose to divide 12j by k+1

  6. anonymous
    • 5 years ago
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    I divided 12j by k-1, not k+1. We have to do that as a means of linking something from statement S(k) to statement S(k+1). When we compare statements S(k) and S(k+1), they have the following factors in common: \[k(k+1)(k+2)\] and we want to try and find a way of incorporating the 12j from S(k) into S(k+1). It just turns out that by looking at S(k), if we divide through by (k-1), we get k(k+1)(k+2) = 12j/(k-1). We can now substitute 12j/(k-1) for k(k+1)(k+2) in S(k+1) and we get the final statement. If this sounds a little bizarre, I apologize - it's difficult to convey everything that's needed over this site. Ask again if you need further clarification and I'll try my best.

  7. anonymous
    • 5 years ago
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    I am understand everything you are telling me it is just I don't usually deal with dividing something over... what concerns me is that when we divide by n-1 we have \[12(k(n+3)/(n+1))\] which may not be an integer and woudn't follow the definition of divisibility

  8. anonymous
    • 5 years ago
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    Ah, okay, I take your point. I'm not on all cylinders today - off sick. I'll take another look.

  9. anonymous
    • 5 years ago
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    thanks I appreciate it :)

  10. anonymous
    • 5 years ago
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    Sorry nxg927, I had some things to do. I had another look. I think one way of going about it is by proving that your expression is divisible by both 3 and 4 (by induction), and therefore by 12. I've just sketched out something rough: 1. First prove 3 divides three consecutive numbers and then use that fact. Prove 3|k(k+1)(k+2). S(k): 3p = k(k+1)(k+2) for some integer p. S(k+1): (k+1)(k+2)(k+3) = k(k+1)(k+2)+3(k+1)(k+2) (expanding the last factor) = 3p + 3(k+1)(k+2) = 3m, for some integer m. Now going back to your expression. For S(k), assume k(k-1)(k+1)(k+2) = 3j for some integer j (that is, we're going to assume your first general case is divisible by 3). Then for S(k+1), (k+1)k(k+2)(k+3) = k(k+1)(k+2) + 3(k(k+1)(k+2)) = k(3M)+3(3M) = 3N for some integers M and N. So, if your test case is divisible by three, then your next case will be too. 2. Prove 4 divides your expression. S(k): 4p = (k^3-k)(k+2) = k(k^3+2k^2-k-2). S(k+1) = ((k+1)^3-(k+1))((k+1)+2) = k(k^3+2k-k-2)+4k^2+12k+8 = 4p + 4(k^2+3k+2) =4m, for some integer m. Since your expression is divisible by 3 and 4, it's divisible by 12. I don't know if you'd have to prove that too. I haven't double-checked this, so there may be mistakes. Good luck.

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