Question
Using binomial theorem, prove that $3^{2\text{n}+2}-8\text{n}-9$ is divisible by 64 where $\text{n}\in\text{N}.$
Using binomial theorem, prove that $3^{2\text{n}+2}-8\text{n}-9$ is divisible by 64 where $\text{n}\in\text{N}.$
$3^{2\text{n}+2}-8\text{n}-9$
$=3^{2(\text{n}+1)}-8\text{n}-9$
$=9^{\text{n}+1}-8\text{n}-9$
$=(1+8)^{\text{n}+1}-8\text{n}-9$
$\Big({^{\text{n}+1}}\text{C}_0+{^{\text{n}+1}}\text{C}_18^1+{^{\text{n}+1}}\text{C}_28^2+.....+{^{\text{n}+1}}\text{C}_{\text{n}+1}8^{\text{n}+1}\Big)-8\text{n}-9$
$\Big(1+8(\text{n}+1)64^{\text{n}+1}\text{C}_2+64(8)^{\text{n}-1}\Big)-8\text{n}-9$
$=64\Big({^{\text{n}+1}\text{C}}_2+......+8^{\text{n}-1}\Big)$
Thus, $3^{2\text{n}+2}-8\text{n}-9$ is divisible by 64.
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