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Homework Help: Prove product of 3 consecutive naturals is even

  1. Dec 4, 2012 #1
    1. Prove product of 3 consecutive naturals is even data

    i just want to know if this proof is acceptable, or is there a simpler proof.

    3. The attempt at a solution

    Proof: suppose if the product 3 consecutive naturals is dividable by 2 then it is even.

    Base case: n = 1

    n (n + 1) (n + 2 )
    since 6 is dividable by 2 base case holds

    Inductive Hypo If n holds then (n+1)

    inductive step

    (n + 1) (n + 2)(n+3)
    n^3 + 6n^2 + 11n + 6

    n(n^2 + 11)+ 6 (n^2 +1)

    since n (n^2 + 11) is even and 6 (n^2 + 1) is even,

    and even + even = even.

    since all even numbers are divisible by 2, thus n +1 hold.

    hence the product of 3 consecutive naturals is even.
    QED(i know that some sentences are missing but that's how i did the proof)
  2. jcsd
  3. Dec 4, 2012 #2


    Staff: Mentor

    Where is your induction hypothesis? You will need it in the inductive step.
    Why is n(n2 + 11) even? You can't just say this.
    There's a much simpler way to prove that n(n + 1)(n + 2) is even. All you need are two cases:
    1. Assume that n is even.
    2. Assume that n is odd.
  4. Dec 4, 2012 #3
    Ohh lol i thought about it, but it felt to simple so didn't do that.

    Why is n(n2 + 11) even?

    Because if n is odd then n^2 + 11 is even and even multiple by odd is even.
    if n is even then even multiple by odd is even. I should add that but, now just gone do what you said it easier.
  5. Dec 4, 2012 #4


    User Avatar
    Homework Helper

    Then explicitly show this.
    If n is odd, thus n = 2k+1 then

    [tex]n(n^2+11) = (2k+1)((2k+1)^2+11)[/tex]




    Which is even... etc.
  6. Dec 5, 2012 #5


    User Avatar
    Science Advisor
    Homework Helper

    The simplest proof would require the solver to see that a sufficient condition for the product of 3 terms to be even is that the simplest subproduct be even, i.e. n(n+1).
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