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Question
Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.
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Solution
We have to prove that 9n + 1 - 8n - 9 = 64k
∴ 9 n+1 - 8n - 9 = (8 + 1)n+ 1 - 8n - 9 [put 9 = 8 + 1]
= \[\ce{[^{n + 1}C_0 8^{n + 1} + ... + ^{n + 1}C_{n-2} 8^3 + ^{n + 1} C_{n - 1} 8^2 + ^{n + 1} C_n 8 + ^{n + 1}C_{n + 1}] - 8n - 9}\]
= \[\ce{^{n + 1} C_0 8 ^{n + 1} + ... + ^{n + 1}C_{n - 2} 8^3 + ^{n + 1}C_{n - 1}8^2 + (n + 1) 8 + 1 - 8n - 9}\]
= \[\ce{^{n + 1}C_0 8 ^{n + 1} + ... + ^{n + 1}C_{n - 2}8^3 + ^{n + 1}C_{n - 1}8^2 + 8n + 8 + 1 - 8n - 9}\]
= \[\ce{^{n + 1}C_0 8^{n + 1} + ... + ^{n + 1}C_{n - 2}8^3 + ^{n + 1}C_{n - 1}8^2}\]
= \[\ce{8^2 [^{n + 1}C_0 8^{n - 1} + ... + ^{n + 1}C_{n - 2} 8 + ^{n + 1}C_{n - 1}]}\]
= 64k [where, k = n + 1C0 8n - 1 + .... + n + 1Cn - 1]
Hence, 9n + 1 - 8n - 9 is divisible by 64, whenever n is a positive integer.
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