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Thread: Mathematical Induction

  1. #1
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    Mathematical Induction

    Question:

    Use the principal of Mathematical Induction to prove 2|(n2+n) for all n>=0
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by oldguy View Post
    Question:

    Use the principal of Mathematical Induction to prove 2|(n2+n) for all n>=0
    for n=0, $\displaystyle 2|(0^2 + 0) = 2|0$

    suppose it is true that for n=k, $\displaystyle 2|(k^2 + k)$.
    show that if n=k+1, then $\displaystyle 2|((k+1)^2 + (k+1))$.

    now, $\displaystyle (k+1)^2 + (k+1) = k^2 + 2k + 1 + k + 1 = k^2 + k + 2k + 2 = k^2 + k + 2(k + 1)$..
    notice that, $\displaystyle 2|k^2 + k$ and $\displaystyle 2|2(k + 1)$

    therefore, $\displaystyle 2|k^2 + k + 2(k + 1)$ or $\displaystyle 2|(k+1)^2 + (k+1)$ . QED
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  3. #3
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    Hello, oldguy!

    Use Mathematical Induction to prove: .$\displaystyle 2\,|\,(n^2+n)$ for all $\displaystyle n \geq 0$
    $\displaystyle S(n)\!: \;n^2+n$ is a multiple of 2.


    Verify $\displaystyle S(1)\!:\;\;1^2 + 1 \:=\:2$ . . . True!


    Assume $\displaystyle S(k)\!:\;\;k^2 + k \:=\:2a$ for some integer $\displaystyle a$


    Add $\displaystyle 2k+2$ to both sides: .$\displaystyle k^2 + k + {\color{blue}2k + 2} \;=\;2a + {\color{blue}2k + 2}$

    We have: .$\displaystyle k^2 + 2k + 1 + k + 1 \;=\;2a + 2k + 2$

    . . . . . . . .. $\displaystyle \underbrace{(k+1)^2 + (k+1)}_{\text{Left side of S(k+1)}} \;=\;\underbrace{2(a + k + 1)}_{\text{multiple of 2}} $


    Therefore, we have proved $\displaystyle S(k+1).$
    . . The inductive proof is complete.

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  4. #4
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    Thanks for the help.
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