$\forall n \in N; 2^{3 n} - 1$ is always divisible by

7

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9

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8

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3

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Solution

The correct option is B $7$
$S = 2^{3 n} - 1 = {(1 + 7)}^{n} - 1 = 1 +^{7} C_{1} 7 +^{7} C_{2} 7^{2} + . . . +^{7} C_{n} 7^{n} - 1$
$\Rightarrow S = 7 λ$
Therefore, $S$ is divisible by $7$
Ans: A

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