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Math Help - Induction

  1. #1
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    Induction

    Use induction to show that 2|n(n+1)

    n=1, \ 2|2

    Assume p(k) is true.

    2|(k+1)(k+2)\Rightarrow 2|(k^2+3k+2)

    2m=k(k+1)

    Now what?
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  2. #2
    Senior Member eumyang's Avatar
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    Assume n = k is true.
    \Rightarrow 2|(k)(k+1) \Rightarrow k(k + 1) = 2m

    Show true for n = k + 1:
    \begin{aligned}<br />
(k + 1)(k + 2) &= k(k + 1) + 2(k + 1) \\<br />
&= 2m + 2(k + 1) \\<br />
&= 2(m + k + 1) \\<br />
&\Rightarrow 2|(k + 1)(k + 2)<br />
\end{aligned}
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  3. #3
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    Quote Originally Posted by dwsmith View Post
    Use induction to show that 2|n(n+1)

    n=1, \ 2|2

    Assume p(k) is true.

    2|(k+1)(k+2)\Rightarrow 2|(k^2+3k+2)

    2m=k(k+1)

    Now what?
    K or k+1 is even.
    Any amount of an even value is even.

    Via induction...

    P(k)

    k(k+1) is even ?

    P(k+1)

    (k+1)(k+2) is even ?

    Try to prove that P(k) being true also causes P(k+1) to be true

    Proof

    (k+2)(k+1)=k(k+1)+2(k+1)

    If P(k) is true, then P(k+1) is true as 2(k+1) is a multiple of 2 and so is even.
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