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$\left(2^{3 \mathrm{n}}-1\right)$ will be divisible by $(\forall \mathrm{n} \in \mathrm{N})$
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Verified Answer
The correct answer is:
7
$$
\begin{aligned}
& \text { Hints: } 2^{3 \mathrm{n}}=(8)^{\mathrm{n}}=(1+7)^{\mathrm{n}}=={ }^{\mathrm{n}} \mathrm{C}_0+{ }^{\mathrm{n}} \mathrm{C}_1 7+{ }^{\mathrm{n}} \mathrm{C}_2 7^2+\ldots \ldots .+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} 7^{\mathrm{n}} \\
& \Rightarrow 2^{3 \mathrm{n}}-1=7\left[{ }^{\mathrm{n}} \mathrm{C}_1+{ }^{\mathrm{n}} \mathrm{C}_2 7+\ldots \ldots \ldots \ldots+{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{n}} 7^{\mathrm{n}-1}\right] \\
& \therefore \text { divisible by } 7
\end{aligned}
$$
\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}
$$
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