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Math Help - Proof by induction

  1. #1
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    Proof by induction

    Can anyone help me get started with this proof using induction? Thank a lot everyone.

    Prove that the product of any three consecutive positive integers is divisible by 6.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jzellt View Post
    Can anyone help me get started with this proof using induction? Thank a lot everyone.

    Prove that the product of any three consecutive positive integers is divisible by 6.

    Base case 1 \times 2 \times 3 = 6 so the proposition is true for the consecutive numbers 1,\ 2,\ 3.

    Suppose the proposition is true for three consecutive numbers starting from k.

    Then 6 divides [k \times (k+1) \times (k+2)] , now consider the product of three consecutive integers starting for k+1:

    (k+1) \times (k+2) \times (k+3) = k \times (k+1) \times (k+2) + 3 \times (k+1) \times (k+2)

    but 6 divides [k \times (k+1) \times (k+2)] and as one of (k+1) and (k+2) must be even 6 divides [3 \times (k+1) \times (k+2)] , and so we conclude that 6 divides [(k+1) \times (k+2) \times (k+3)] .

    Now we have proven the base case and that if the proposition is true for three consecutive integers starting from k it is true for three consecutive integers starting from k+1, and so by induction it is true for every set of
    three consective positive integers.

    RonL
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  3. #3
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    Hello, jzellt!

    A slight variation of CaptainBlack's proof.


    Prove that the product of any three consecutive positive integers is divisible by 6.
    We want to prove

    . . \text{For all }n\in I^+,\;n(n+1)(n+2) \:=\:6a\;\;\text{for some integer }a.


    Verify S(1)\!:\;\;1\cdot2\cdot3\:=\:6 . . . True!


    Assume S(k)\!:\;\;k(k+1)(k+2) \;=\;6a for some integer a


    Add 3(k+1)(k+2) to both sides:
    . . k(k+1)(k+2) \:{\color{blue}+\:3(k+1)(k+2)} \;=\;6a \:{\color{blue}+ \:3(k+1)(k+2)}


    The left side is: . k(k+1)(k+2)+ 3(k+1)(k+2)
    . . Factor: . (k+1)(k+2)(k+3)\;\;{\color{red}\text{[L]}}


    The right side is: . 3[2a + (k+1)(k+2)]
    . . (k+1)(k+2) is the product of two consecutive integers.
    . . Hence, it is divisible by 2:. (k+1)(k+2) \:=\:2b
    The right side is:. 3(2a+2b) \:=\:6(a+b)\;\;{\color{red}\text{[R]}}


    From [L] and [R],we have: . (k+1)(k+2)(k+3)\:=\:6(a+b)

    We have established S(k+1)..The inductive proof is complete.

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