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Mathematical Induction

  1. Oct 11, 2006 #1
    Proposition: 1*2*3+2*3*4+3*4*5+...+n(n+1)(n+2) = [n(n+1)(n+2)(n+3)]/4

    Step (1): If n=1 then LHS (left hand side) = 6, and RHS = 6
    Thus, P1 is true.

    Step (2): If Pk is true then
    k(k+1)(k+2) = [k(k+1)(k+2)(k+3)]/4

    k(k+1)(k+2) + [k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + [k+1]([k+1]+1)([k+1]+2)

    k(k+1)(k+2) +[k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + (k+1)(k+2)(k+3)

    [k(k+1)(k+2)(k+3) + 4(k+1)(k+2)(k+3)]/4 ---> Common denominator - 4.

    Then...what do i do? any clues? I have not proven my proposition, yet.
    Last edited: Oct 11, 2006
  2. jcsd
  3. Oct 11, 2006 #2


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    Think about this equality. Does it make sense?
  4. Oct 11, 2006 #3
    oopsies i meant to write
    k(k+1)(k+2) + [k+1]([k+1]+1)([k+1]+2) = [k(k+1)(k+2)(k+3)]/4 + [k+1]([k+1]+1)([k+1]+2)
  5. Oct 11, 2006 #4


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    Right, and that implies [tex]k(k+1)(k+2) + (k+1)(k+2)(k+3) = \frac{k(k+1)(k+2)(k+3)+4(k+1)(k+2)(k+3)}{4}[/tex]. The rest is pretty obvious.
  6. Oct 11, 2006 #5
    im not very good at factoring, but ill give it a try and you see whether im on the right track:
    let k+1= A, k+2= B, k+3= C
    k(A)(B)(C) + 4(A)(B)(C)/4
    am i on the right track?
    Then?????? How do i prove that P(k+1) is true whenever Pk and P1 are true???!
    Last edited: Oct 11, 2006
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