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Math Help - 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 7^{n}-1=6p

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

    7^{n}-1=6k

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

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

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

    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 7^n-1 is divisible by 6, n>0

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

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


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

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

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

    . . . . . . . . . . . 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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