Using binomial theorem, prove that 23n – 7n – 1 is divisible by 49, where n ∈ N.
Proof: We can write23n – 7n – 1 as
8n - 7n - 1 - - - - - (1)
Now,
8n = (1 + 7)n
And we know that,
(x + a)n = nC0 .(x)n(a)0 + nC1 .(x)n - 1(a)1 + nC2 .(x)n - 2(a)2 + … + nCn .(x)0(3y)n
Similarly,
(1 + 7)n = nC0 + nC171 + nC2.72 + nC3.73+ … + nCn.7n
(1 + 7)n = 8n = 1 + 7n + 49[nC2 + nC3.71+ … + nCn.7n - 2]
(1 + 7)n = 8n - 1 - 7n = 49 × (Any integer)
Now,
= 8n - 1 - 7n is divisible by 49
Hence, 23n - 7n - 1 is divisible by 49
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