Using binomial theorem, prove that (23n - 7n -1) is divisible by 49, where n N.

To prove: (23n - 7n -1) is divisible by 49, where n N


Formula used: (a+b)n = nC0an + nC1an-1b + nC2an-2b2 + …… +nCn-1abn-1 + nCnbn


(23n - 7n -1) = (23)n – 7n – 1


8n - 7n – 1


(1+7)n – 7n – 1


nC01n + nC11n-17 + nC21n-272 + …… +nCn-17n-1 + nCn7n – 7n – 1


nC0 + nC17 + nC272 + …… +nCn-17n-1 + nCn7n – 7n – 1


1 + 7n + 72[nC2 + nC37 + … + nCn-1 7n-3 + nCn 7n-2] -7n -1


72[nC2 + nC37 + … + nCn-1 7n-3 + nCn 7n-2]


49[nC2 + nC37 + … + nCn-1 7n-3 + nCn 7n-2]


49K, where K = (nC2 + nC37 + … + nCn-1 7n-3 + nCn7n-2)


Now, (23n - 7n -1) = 49K


Therefore (23n - 7n -1) is divisible by 49


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