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Math Help - Proof by Induction

  1. #1
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    Proof by Induction

    Prove 24|2*7^n+3*5^n-5


    I showed true for n=1:
    2*7+3*5-5=14+15-5=24
    Thus divisible by 24
    Assume true for all k>=1. Need to prove true for n=k+1
    Then 2*7^(k+1)+3*5^(k+1)-5=14*7^k+15*5^k-5
    I get this far and now I get stuck simplifying enough to show it's true.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by kathrynmath View Post
    Prove 24|2*7^n+3*5^n-5


    I showed true for n=1:
    2*7+3*5-5=14+15-5=24
    Thus divisible by 24
    Assume true for all k>=1. Need to prove true for n=k+1
    Then 2*7^(k+1)+3*5^(k+1)-5=14*7^k+15*5^k-5
    I get this far and now I get stuck simplifying enough to show it's true.
     14\cdot7^k+15\cdot5^k-5 = 12\cdot\left(7^k+5^k\right)+\left(2\cdot7^k+3\cdot  5^k-5\right)

    Now note that  7^k+5^k is even.
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  3. #3
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    Quote Originally Posted by kathrynmath View Post
    Prove 24|2*7^n+3*5^n-5


    I showed true for n=1:
    2*7+3*5-5=14+15-5=24
    Thus divisible by 24
    Assume true for all k>=1. Need to prove true for n=k+1
    Then 2*7^(k+1)+3*5^(k+1)-5=14*7^k+15*5^k-5
    I get this far and now I get stuck simplifying enough to show it's true.

    2\cdot 7^{k+1}+3\cdot 5^{k+1}-5=[2\cdot 7^k+3\cdot 5^k-5]+12[7^k+5^k] ,so the first parentheses

    above is divisible by 24 by the inducive hypothesis, and the second parentheses is obviously even, so...

    Tonio
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