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sasogeek
 3 years ago
prove that \( 7^n 1\) is divisible by 6 for all n\(\ge \) 1
sasogeek
 3 years ago
prove that \( 7^n 1\) is divisible by 6 for all n\(\ge \) 1

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(7)^n 1 = (6+1)^n 1 Now, use binomial coefficient to open (6+1)^n

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(6+1)^n = nC0 (6)^0 * 1^n + nC1 * 6^1 + 1^(n1)+...+nCn * 6^6*1^0 =1 + nC1*6^1*1^(n1)+...+nCn*6^n *1^0 This gives, (6+1)^n1 = nC1*6^1*1^(n1)+...+nCn*6^n *1^0

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Now, u can take 6 common from all the terms of (7)^n1. Thus, it is divisible by 6 for all n>=1

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0just 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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