Q53 / 160 · WBJEE 2010 · Mathematics

$\left(2^{3 \mathrm{n}}-1\right)$ will be divisible by $(\forall \mathrm{n} \in \mathrm{N})$

  • A
    25
  • B
    8
  • Correct answer
    C
    7
  • D
    3

Solution

$$ \begin{aligned} & \text { Hints: } 2^{3 \mathrm{n}}=(8)^{\mathrm{n}}=(1+7)^{\mathrm{n}}=={ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1 7+{ }^{\mathrm{n}} \mathrm{C}_2 7^2+\ldots \ldots .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} 7^{\mathrm{n}} \\ & \Rightarrow 2^{3 \mathrm{n}}-1=7\left[{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2 7+\ldots \ldots \ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} 7^{\mathrm{n}-1}\right] \\ & \therefore \text { divisible by } 7 \end{aligned} $$

Hence, the correct answer is option: 'C'.

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