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# Show that $9^{n+1} – 8n – 9$ is divisible by $64$, whenever $n$ is a positive integer.

• $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n, where b^0 = 1 = a^{n-n} 9^{n+1} = (1+8)^{n+1} \rightarrow which can be evaluated using binomial theorem as follows: 9^{n+1} = (1+8)^{n+1} = \;^{n+1}\large C$$_0\;(1)^{n+1}8^0 + \;^{n+1}\large C$$_0\;(1)^{n+1-1}8^1 + \;^{n+1}\large C$$_2\;(1)^{n+1-2}8^2+...+ \;^{n+1}\large C$$_{n+1}\;(1)^{n+1-n-1}8^{n+1} \qquad = 1 + (n+1) 8 + ^{n+1}\large C$$_2 8^2 + ^{n+1}\large C$$_3 8^3 +.... \qquad = 9 + 8n + 64 ( ^{n+1}\large C$$_2 + ^{n+1}\large C$$_3 8 +....)$
$\qquad = 9+ 8n + 64k$ where k is a natural number as the series evaluates to a positive natural number.
$\Rightarrow 9^{n+1} = 9 + 8n + 64k$
$\Rightarrow 9^{n+1} - 8n - 9 = 64k \rightarrow$ which is divisible by $64$ for all $n \in \text{Positive integers}$