Question

Show that \(9^{ n+1} -8n -9\) is divisible by \(64\), whenever n is a positive integer.

Answer 1

In order to show that \(9^{n+1}-8n-9 \) is divisible by \(64\), it has to be proved that,
\(9^{n+1}-8n-9 \)=\(64k\), where k is some natural number
By Binomial Theorem,
\((1+a)^m\) = \(^mC_0+^mC_1a+ ^mC_2 a^2+...+ ^mC_m a^m\)
For \(a = 8\) and \(m = n+ 1\), we obtain
\((1+8)^{n+1}\) = \(^{n+1}C_0+ ^{n+1}C_1(8) + ^{n+1}C_2 (8)^2 + ... + ^{n+1}C_{n+1}(8n+1)\)
⇒ \( 9^{n+1} = 1 + (n+1)(8) + 8^2 [^{n+1}C_2 + ^{n+1}C_3×8+...+ ^{n+1}C_{n+1} (8)^{n-1}]\)
⇒ \( 9^{n+1} = 9+8n +64[ ^{n+1}C_2 + ^{n+1}C_3 × 8+...+^{n+1}C_{n+1} (8)^{n-1}]\)
⇒ \(9^{n+1}-8n-9=64k\), where \(k \) =  \(^{n+1}C_2 + ^{n+1}C_3 × 8+...+^{n+1}C_{n+1} (8)^{n-1}\) is a natural number.

Thus, \(9^{n+1}-8n-9\) is divisible by \(64\), whenever \(n\) is a positive integer.