Question
Using binomial theorem, prove that $2^{3\text{n}}-7\text{n}-1$ is divisible by 49 where $\text{n}\in\text{N}.$
Using binomial theorem, prove that $2^{3\text{n}}-7\text{n}-1$ is divisible by 49 where $\text{n}\in\text{N}.$
$2^{3\text{n}}-7\text{n}-1$
$=2^{3(\text{n})}-7(\text{n})-1$
$=8^\text{n}-7\text{n}-1$
$=(1+7)^\text{n}-7\text{n}-1$
$==({^\text{n}\text{C}}_0+{^\text{n}\text{C}}_1(7)^1+{^\text{n}\text{C}}_2(7)^2+....{^\text{n}\text{C}}_\text{n}(7)^\text{n})-7\text{n}-1$
$=(1+7\text{n}+49^\text{n}\text{C}_2+......49(7)^{\text{n}-2})-7\text{n}-1$
$=49({^\text{n}\text{C}}_2+......+7^{\text{n}-2})$
$\therefore2^{3\text{n}}-7\text{n}-1$ is divisible by 49
Hence, proved
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