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Thread: Mathematical Induction

  1. #1
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    Question Mathematical Induction

    Plz help on these:

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


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

    Thnx in advance.
    Last edited by cu4mail; Jan 7th 2008 at 10:00 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
    $\displaystyle 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:

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

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

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

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

    now continue
    Last edited by Jhevon; Jan 7th 2008 at 10:36 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 $\displaystyle n^2 > 2n +1 $ for $\displaystyle 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:

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

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

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

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

    $\displaystyle > 2k + 1 + 2k + 1$ .....since $\displaystyle k^2 > 2k + 1$

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

    $\displaystyle = 2k + 8$

    $\displaystyle > 2k + 3$

    $\displaystyle = 2(k + 1) + 1$

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

    Thus we have proven $\displaystyle P(n)$ by induction
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