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Show that 9^(n+1)-8n-9is divisible by 64...

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

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The correct Answer is:
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`(1+8)^(n+1)-8n-9`
`=overset(n+1)underset(r=0)(Sigma).^(n+1)C_(r).(1)^(n+1-r).8^(r)-8n-9`
`[.^(n|1)C_(0).8^(0)+^(n+1)C_(1),8^(1)+^(n+1)C_(2).8^(2)`
`+^(n+1)C_(3).8^(3)+......+^(n+1)C_(n+1).8^(n+1)]-8n-9`
`[1+8.(n+1)+^(n+1)C_(2).8^(2)+^(n+1)C_(3).8^(3)`
`+...+^(n+1)C_(n+1),8^(n+1)]-8n-9`
`=^(n+1)C_(2).8^(2)+^(n+1)C_(3).8^(3)+......+^(n+1)C_(n+1).8^(n+1)`
`=8^(2[.^(n+1)C_(2)+^(n+1)C_(3),8+....+^(n+1)C_(n+1),8^(n-1)]`
Which is divisible by 64
Therefore `(9^(n+1)-8n-9)` is divisible by 64
Hence Proved.
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