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