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sasogeek

  • 2 years ago

prove that \( 7^n -1\) is divisible by 6 for all n\(\ge \) 1

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  1. Yahoo!
    • 2 years ago
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    Using PMI ....

  2. sauravshakya
    • 2 years ago
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    (7)^n -1 = (6+1)^n -1 Now, use binomial coefficient to open (6+1)^n

  3. sauravshakya
    • 2 years ago
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    (6+1)^n = nC0 (6)^0 * 1^n + nC1 * 6^1 + 1^(n-1)+...+nCn * 6^6*1^0 =1 + nC1*6^1*1^(n-1)+...+nCn*6^n *1^0 This gives, (6+1)^n-1 = nC1*6^1*1^(n-1)+...+nCn*6^n *1^0

  4. sauravshakya
    • 2 years ago
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    Now, u can take 6 common from all the terms of (7)^n-1. Thus, it is divisible by 6 for all n>=1

  5. Coolsector
    • 2 years ago
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    just add another way - by induction check for n =1 we get 6/6 check for n = 2 we get 48/6 = 8 now lets assume that it is correct for n =k now check for n= k+1 [7^(k+1) + 1] / 6 = (7*7^k + 1 )/6 = (6*7^k + 7^k + 1 )/6 = 6*7^k/6 + (7^k + 1)/6 the first term becomes 7^k and the second by is divisible by our assumption.

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