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JRHarrison90

  • 3 years ago

Use mathematical induction to prove that for all positive integers 3 is a factor of n^3 + 2n.

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  1. anonymous
    • 3 years ago
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    wow induction for this?

  2. anonymous
    • 3 years ago
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    first check that it is true if \(n=1\) which it is because \(1^2+2\times 1=3\) and 3 is a factor of 3

  3. anonymous
    • 3 years ago
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    now assume that it is true if \(n=k\) meaning assume that for all \(k\) you have \[k^3+2k\] is divisible by 3 now lets see if we can prove it is true for \(k+1\)

  4. anonymous
    • 3 years ago
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    that is, see if you can show that \[(k+1)^3+2(k+1)\] is divisible by 3, if you get to assume that \(k^3+2k\) is mostly it is algebra to see if you can arrange to get a \(k^3+2k\) out of \[(k+1)^3+2(k+1)\]

  5. anonymous
    • 3 years ago
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    but i get the feeling i am talking to myself, so i will be quiet now

  6. electrokid
    • 3 years ago
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    from step2, \[k^3+2k=3a\\ k(k^2+2)=3a\] \[ (k+1)^3+2(k+1)=(k+1)[(k+1)^2+2]\\ \qquad=(k+1)[k^2+2k+1+2]\\ \qquad=(k+1)\left[{3a\over k}+2k+1\right]\\ \qquad=3a+2k^2+k+{3a\over k}+2k+1\\ \qquad=(3a)\left(1+{1\over k}\right)+3k+(2k^2+1) \] each of the three terms is a factor of "3"

  7. electrokid
    • 3 years ago
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    proof for \[2k^2+1=2\left({3a\over2}-2\right)+1=3a\]

  8. JRHarrison90
    • 3 years ago
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    Makes sense.

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