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Math Help - Induction divisibility P2

  1. #1
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    Induction divisibility P2

    Could somebody help me with these questions? I just need to prove that n=k+1 using the assumption for n=k

    Prove:
    1) 7^2n - 48n -1 for 2304 for n=1,2,3...
    2) (3n+1).7^n -1 is divisible by 9 for n=1,2,3...
    3) 10^n +3[4^(n+2)] +5 is divisible by 9 for n=1,2,3...

    Thank you !
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  2. #2
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    <br />
(3n + 1)\cdot 7^n - 1 = 9k; k \in \mathbb{N}<br />
    basis:
    <br />
(3(1) + 1)\cdot 7^{(1)} - 1 = 9k<br />
    <br />
27 = 9k<br />
    <br />
k = 3<br />

    hypothesis:
    <br />
(3n + 1)\cdot 7^n - 1 = 9k; k \in \mathbb{N}<br />
    <br />
(3(n + 1) + 1)\cdot 7^{n + 1} - 1 = 9k; k \in \mathbb{N}<br />
    <br />
7\cdot 7^n(3n + 1 + 3)  - 1 = 9k; k \in \mathbb{N}<br />
    <br />
7\cdot 7^n(3n + 1) - 1 + 7\cdot 7^n\cdot 3 = 9k; k \in \mathbb{N}<br />

    finished. We have 7 times a multiple of 9 and 7 times a multiple of 63, which is also divisble by 9.
    * actually there is an error here- can someone do n = 2 for me?

    for 3)
    <br />
9\cdot 10^n + 10^n + 9\cdot 4^{m + 2} + 3\cdot4^{n + 2} + 5 = 9k<br />
    Last edited by bmp05; February 6th 2010 at 08:56 AM.
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  3. #3
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    Quote Originally Posted by christina View Post
    Could somebody help me with these questions? I just need to prove that n=k+1 using the assumption for n=k

    Prove:
    1) 7^2n - 48n -1 for 2304 for n=1,2,3... can you resubmit this?
    2) (3n+1).7^n -1 is divisible by 9 for n=1,2,3...
    3) 10^n +3[4^(n+2)] +5 is divisible by 9 for n=1,2,3...

    Thank you !
    2)

    \frac{(3n+1)7^n-1}{9}=natural\ number

    \frac{[3(n+1)+1]7^{n+1}-1}{9}=\frac{\left([3n+1+3]7^n\right)7-1}{9}

    =\frac{\left((3n+1)7^n\right)7+3(7)7^n-1}{9}

    =\frac{(3n+1)7^n-1+(3n+1)7^n(6)+7(3)7^n}{9}

    =\frac{(3n+1)7^n-1+(18n)7^n+(6)7^n+(21)7^n}{9}

    =\frac{(3n+1)7^n-1+(18n)7^n+(27)7^n}{9}

    As the other 2 terms of the numerator are divisible by 9,
    then if the first term is divisible by 9, the hypothesis is proven if it's true for n=1.


    3.

    \frac{10^n+3\left(4^{n+2}\right)+5}{9}=natural\ number

    \frac{10^{n+1}+3\left(4^{n+3}\right)+5}{9}=\frac{(  10)10^n+3\left(4^{n+2}\right)+(9)10^n+(3)3\left(4^  {n+2}\right)+5}{9}

    =\frac{\left(10^n+3\left(4^{n+2}\right)+5\right)+9  \left(10^n+4^{n+2}\right)}{9}

    Therefore, if

    10^n+3\left(4^{n+2}\right)+5

    is divisible by 9

    10^{n+1}+3\left(4^{n+3}\right)+5

    is also divisible by 9.

    Hence check foe n=1.
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