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Math Help - proof by induction

  1. #1
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    proof by induction

    prove by induction that 4^n-1 is divisible by three for all posotive integers of n
    any help would be appriciated thanks.
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  2. #2
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    Hi

    Initialization of induction is easy.

    Let's suppose that for a given n 4^{n}-1 is divisible by 3.
    You need to demonstrate that 4^{n+1}-1 is divisible by 3.
    Hint : 4^{n+1}-1 = 4(4^{n}-1)+3.
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  3. #3
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    prove 4^n - 1 is divisible by 3

    true for n = 1

    assume true for n, show true for (n+1) ...

    since 4^n - 1 is divisible by 3, then 4^n - 1 = 3k , k a positive integer.

    4^n - 1 = 3k

    4(4^n - 1) = 4(3k)

    4^{n+1} - 4 = 12k

    4^{n+1} - 1 - 3 = 12k

    4^{n+1} - 1 = 12k + 3

    4^{n+1} - 1 = 3(4k + 1)

    since (4k+1) is also a positive integer, 4^{n+1} - 1 is a multiple of 3, and therefore, is divisible by 3.
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  4. #4
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    Hello, hmmmm!

    Here's another way . . .


    Prove by induction that 3|(4^n-1) for all positive integers n.

    Verify S(1)\!:\;\;4^1-1 \:=\:3 . . . true

    Assume S(k)\!:\;\;4^k-1 \:=\:3a\,\text{ for some integer }a.


    Add 3\!\cdot\!4^k to both sides: . {\color{blue}3\!\cdot\!4^k} + 4^k - 1 \;=\;3a + {\color{blue}3\!\cdot\!4^k}


    Factor: . (3 + 1)4^k - 1  \;=\;3\left(a + 4^k\right)

    . . . . . . . . 4\!\cdot4^k - 1 \;=\;3\left(a + 4^k\right)

    . . . . . . . . 4^{k+1}-1 \;=\;\underbrace{3\left(a + r^k\right)}_{\text{a multiple of 3}}


    We have proved S(k+1) . . . The inductive proof is complete.

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