Question

For each natural number, the statement $P\left(n\right)=2^{3n}-1$ is divisible by

• A. $10$
• B. $6$
• C. $7$
• D. None of these.

Answer 1

Answer: (C)

$7$

$P\left(n\right)=2^{3n}-1=8^n-1$

$P\left(n\right)={(1+7)}^n-1$

$\Rightarrow P\left(n\right)=1{+}^n{C}_1\times 7{+}^n{C}_2\times 7^2+...{+}^n{C}_n\times 7^n-1$

$\Rightarrow P\left(n\right)=7({}^n{C}_1{+}^n{C}_{2}7+...{+}^n{C}_n{7}^{n-1})$

Therefore, $P\left(n\right)$ is divisible by $7$.