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Question
If n is a positive integer, using Binomial theorem, show that, 9n+1 − 8n − 9 is always divisible by 64
Solution
(1 + x)n = nC0 + nC1x + nC2x2 + ........ + nCn−1 xn−1 + nCnxn
Put x = 8 we get
(1 + 8)n = nC0 + nC1(8) + nC2(8)2 + ......... + nCn–1 8n–1 + nCn . 8n
9n = nC0 + nC1(8) + nC2(8)2 + ......... + nCn–1 8n–1 + nCn . 9n = 1 + 8n + nC2 × 82 + ........ +
nCn–1 8n–1 + nCn . 8n
9n - 8n – 1 = nC2 × 82 + ........ + nCn–1 8n–1 + nCn × 8n
9n - 8n – 1 = 82 [nC2 + .......... + nCn–1 × 8n–3 + nCn × 8n–2]
Which is divisible by 64 for all positive integer n.
∴ 9n – 8n – 1 is divisible by 64 for all positive integer n.
Put n = n + 1 we get
9n + 1 – 8 (n + 1) – 1 is divisible by 64 for all possible integer n
(9n + 1 – 8n – 8 – 1) is divisible by 64
∴ 9n + 1 – 8n – 9 is always divisible by 64
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