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Thread: Mathematical induction

  1. #1
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    Mathematical induction

    Prove (7^n)-1 is divisible by 6,n>0
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  2. #2
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    Assume $\displaystyle 7^{n}-1=6p$

    Then show that $\displaystyle 7^{n+1}-1=6k$

    $\displaystyle 7^{n}-1=6k$

    $\displaystyle 7\cdot (7^{n}-1)=7\cdot 6k$

    $\displaystyle 7^{n+1}-7=6\cdot 7k$

    $\displaystyle 7^{n+1}-1=6\cdot 7k+6$

    $\displaystyle 7^{n+1}-1=6(7k+1)$...QED
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  3. #3
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    Hello, Knigh17!

    A variation of galactus' solution . . .


    Prove $\displaystyle 7^n-1$ is divisible by 6, $\displaystyle n>0$

    Verify $\displaystyle S(1)\!:\;\;7^1 - 1 \:=\:6$ . . . true.

    Assume $\displaystyle S(k)\!:\;\;7^k - 1 \:=\:6a\;\;\text{ for some integer }a$


    Add $\displaystyle 6\!\cdot\!7^k$ to both sides: .$\displaystyle 7^k + {\color{blue}6\!\cdot\!7^k} - 1 \;=\;6a + {\color{blue}6\!\cdot\!7^k}$

    We have: .$\displaystyle (1 + 6)\!\cdot\!7^k - 1 \;=\; 6a + 6\!\cdot\!7^k $

    . . . . . . . . . . . $\displaystyle 7\!\cdot\!7^k - 1 \;=\;6a + 6\!\cdot\!7^k$

    . . . . . . . . . . . $\displaystyle 7^{k+1} - 1 \;=\;\underbrace{6(a + 7^k)}_{\text{a multiple of 6}}\quad\hdots\;QED$

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  4. #4
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    Thanks

    Thanks a lot! now i know how this is solved.
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