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

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Solution

We have $9^{n + 1} = (1 + 8)^{n + 1}$

$= (n + 1)_{C_{0}} + (n + 1)_{C_{1}} (8) + (n + 1)_{C_{2}} (8)^{2} + (n + 1)_{C_{3}} (8)^{3} + . . . + (n + 1)_{C_{n + 1}} (8)^{n + 1}$

$= 1 + (n + 1) \times 8 + {(n + 1)}_{C_{2}} (8)^{2} + {(n + 1)}_{C_{3}} (8)^{3} + . . . + {(n + 1)}_{C_{n + 1}} (8)^{n + 1}$

$= 1 + 8 n + 8 + {(n + 1)}_{C_{2}} (8)^{2} + {(n + 1)}_{C_{3}} (8)^{3} + . . . + {(n + 1)}_{C_{n + 1}} (8)^{n + 1}$

$∴ 9^{n + 1} - 8 n - 9$

$= {n + 1}_{C_{2}} (8)^{2} + {n + 1}_{C_{3}} (8)^{3} + . . . + {n + 1}_{C_{n + 1}} (8)^{n + 1}$

$= 64 [{(n + 1)}_{C_{2}} + {(n + 1)}_{C_{3}} .8 + . . . + {(n + 1)}_{C_{n + 1}} . 8^{n + 1}]$

which shows that $9^{n + 1} - 8 n - 9$ is divisible by 64.

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