Results 1 to 6 of 6

Math Help - prove that n^3 + 5n is divisble by 6 using induction

  1. #1
    Junior Member mremwo's Avatar
    Joined
    Oct 2010
    From
    Tampa, FL
    Posts
    53

    prove that n^3 + 5n is divisble by 6 using induction

    So I started doing this proof then realized I hasn't really used induction. I said that n^3 + 5n is equal to n^3 - n + 6n = n(n^2 - 1) + 6n = n(n-1)(n+1) + 6n = (n-1)(n)(n+1) + 6n

    Now I know that 6n is divisible by 6 and the product of 3 consecutive integers is divisible by 6 so that the entire term is divisble by 6 but I don't know how to write a proper proof using that explanation. I know you can do this using induction although im not sure how. Thanks for your help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    A Plied Mathematician
    Joined
    Jun 2010
    From
    CT, USA
    Posts
    6,318
    Thanks
    4
    Awards
    2
    Well, what's your base case going to be?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Bar0n janvdl's Avatar
    Joined
    Apr 2007
    From
    South Africa
    Posts
    1,630
    Thanks
    6
    Quote Originally Posted by mremwo View Post
    So I started doing this proof then realized I hasn't really used induction. I said that n^3 + 5n is equal to n^3 - n + 6n = n(n^2 - 1) + 6n = n(n-1)(n+1) + 6n = (n-1)(n)(n+1) + 6n

    Now I know that 6n is divisible by 6 and the product of 3 consecutive integers is divisible by 6 so that the entire term is divisble by 6 but I don't know how to write a proper proof using that explanation. I know you can do this using induction although im not sure how. Thanks for your help
    Exactly one of every three consecutive integers is divisible by 3. Same goes for 2. (Except that's every second consecutive integer.)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Mar 2010
    From
    Florida
    Posts
    3,093
    Thanks
    5
    Quote Originally Posted by mremwo View Post
    So I started doing this proof then realized I hasn't really used induction. I said that n^3 + 5n is equal to n^3 - n + 6n = n(n^2 - 1) + 6n = n(n-1)(n+1) + 6n = (n-1)(n)(n+1) + 6n

    Now I know that 6n is divisible by 6 and the product of 3 consecutive integers is divisible by 6 so that the entire term is divisble by 6 but I don't know how to write a proper proof using that explanation. I know you can do this using induction although im not sure how. Thanks for your help
    Since this works \forall n\geq 1

    Show P(1) is true.

    Then let n\leq k where k is an arbitrary integer.

    Assume P(k) is true and then prove P(k+1) is true.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Nov 2010
    From
    Staten Island, NY
    Posts
    451
    Thanks
    2
    Mremwo, your proof looks good to me. It's just not a proof by induction.

    Your proof is actually better than an inductive proof because it proves the statement for all integers n. A proof by induction needs a starting point.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by mremwo View Post
    So I started doing this proof then realized I hasn't really used induction. I said that n^3 + 5n is equal to n^3 - n + 6n = n(n^2 - 1) + 6n = n(n-1)(n+1) + 6n = (n-1)(n)(n+1) + 6n

    Now I know that 6n is divisible by 6 and the product of 3 consecutive integers is divisible by 6 so that the entire term is divisble by 6 but I don't know how to write a proper proof using that explanation. I know you can do this using induction although im not sure how. Thanks for your help
    P(k)

    The proposition for n=k is

    k^3+5k is divisible by 6


    P(k+1)

    The proposition for the "next n" is

    (k+1)^3+5(k+1) is divisible by 6.


    You need to prove that the Induction Domino-Effect exists
    by showing that P(k+1) will be true "if" P(k) is true.

    Hence, either write P(k+1) in terms of P(k),
    when we will see if P(k+1) will be true if P(k) is..

    Or work towards P(k+1) from P(k).

    Proof

    (k+1)^3+5(k+1)=k^3+3k^2+3k+1+5k+5

    =\left(k^3+5k\right)+3k^2+3k+6=\left(k^3+5k\right)  +3k(k+1)+6

    Since k and k+1 are a pair of consecutive even and odd terms, 3k(k+1) is a multiple of 6 for k \ge\ 1

    Then we can see if P(k+1) will be true "if" P(k) is.

    Prove for a base case to topple the dominoes.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. n^3+n=m^4 proving n has to be even and divisble by 16
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: September 22nd 2011, 02:45 PM
  2. 5digit numbers divisble by 6
    Posted in the Statistics Forum
    Replies: 2
    Last Post: September 2nd 2011, 09:08 AM
  3. Replies: 10
    Last Post: June 29th 2010, 12:10 PM
  4. prove by induction
    Posted in the Algebra Forum
    Replies: 3
    Last Post: June 19th 2008, 08:09 AM
  5. Prove by induction....
    Posted in the Algebra Forum
    Replies: 1
    Last Post: April 9th 2008, 11:37 AM

Search Tags


/mathhelpforum @mathhelpforum