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Math Help - Mathematical Induction

  1. #1
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    Question Mathematical Induction

    Plz help on these:

    (a) Prove by mathematical induction that
     6^{n+2} + 7^{2_{n+1}} is divisible by 43 for each positive integer n


    (b) Show that  n^2 > 2n +1 for  n \ge 3 using mathematical induction method.

    Thnx in advance.
    Last edited by cu4mail; January 7th 2008 at 11:00 PM. Reason: 2nd question added
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cu4mail View Post
    Plz help on these:

    (a) Prove by mathematical induction that
     6^{n+2} + 7^{2_{n+1}} is divisible by 43 for each positive integer n
    i'll start you off...

    we wish to prove the statement:

    P(n):6^{n + 2} + 7^{2n + 1} is divisible by 43 for all n \in \mathbb{N}.

    P(1) is true, so we have our inductive step.

    Assume then that P(k) is true for k \ge 1, k \in \mathbb{N}, we show that P(k + 1) is true.

    Now, P(k + 1) = 6^{k + 3} + 7^{2k + 3} = 6^{k + 3} + \underbrace{6 \cdot 7^{2k + 1} - 6 \cdot 7^{2k + 1}}_{\mbox{This is zero}} + 7^{2k + 3}
    .
    .
    .

    now continue
    Last edited by Jhevon; January 7th 2008 at 11:36 PM. Reason: typed n when i should have typed k
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cu4mail View Post
    (b) Show that  n^2 > 2n +1 for  n \ge 3 using mathematical induction method.
    this one is not that bad, in fact, i think giving you a hint would give the answer away anyway, so let's just do it.

    we wish to prove the statement:

    P(n): n^2 > 2n + 1 for all n \ge 3, n \in \mathbb{N}

    Now, P(3): 9 > 7 which is true. So the inductive step is complete.

    Assume P(k) is true for some k \ge 3, k \in \mathbb{N}, we show P(k + 1) is true. that is, we want to show (k + 1)^2 > 2(k + 1) + 1

    now, P(k + 1): (k + 1)^2 = k^2 + 2k + 1

    > 2k + 1 + 2k + 1 .....since k^2 > 2k + 1

    \ge 2k + 1 + 6 + 1 ......since k \ge 3

    = 2k + 8

    > 2k + 3

    = 2(k + 1) + 1

    thus we have shown P(k + 1) is true.

    Thus we have proven P(n) by induction
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