$\forall n \in N, 4^{n} - 3 n - 1$ is divisible by

3

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8

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9

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11

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Solution

The correct options are
A $3$
C $9$
$4^{n} - 3 n - 1 = (1 + 3)^{n} - 3 n - 1 = (1 +^{n} C_{1} 3 +^{n} C_{2} 3^{2} + . . . . . . . . .^{n} C_{n} 3^{n}) - 3 n - 1$
$= (1 + 3 n +^{n} C_{2} 3^{2} + . . . . . . . .) - 3 n - 1 =^{n} C_{2} 3^{2} +^{n} C_{3} 3^{3} + . . . . . . . . . . . . +^{n} C_{n} 3^{n}$
$= 3^{2} (^{n} C_{2} +^{n} C_{3} 3 + . . . . . . . . . . . +^{n} C_{n} 3^{n - 2}) = 9 \times$ Integer.
Hence given number is divisible by $9$ . Hence it is also divisible by $3$ .

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