1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    Mark44

    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

    Mentallic

    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]

    [tex]=(2k+1)(4k^2+4k+1+11)[/tex]

    [tex]=(2k+1)(4k^2+4k+12)[/tex]

    [tex]=2(2k+1)(2k^2+2k+6)[/tex]

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

    dextercioby

    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).
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Prove product of 3 consecutive naturals is even
  1. Prove that k is even (Replies: 4)

Loading...