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

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

Discrete Math
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Using PMI ....
(7)^n -1 = (6+1)^n -1 Now, use binomial coefficient to open (6+1)^n
(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

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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
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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