Question

$7$ is a factor of $2^{3n}-1$ for all natural numbers n.

• A. True
• B. False

Answer 1

The correct option is A True

Let P(n): 7 is a factor of $2^{3n}-1$ be the given statement
Step 1: When $n=1$
$2^{3\left(1\right)}-1=7$ and 7 is a factor of itself.
$\therefore$ P(n) is true for $n=1$
Step 2: Let P(n) be true for $n=k$
$⟹$ 7 is a factor of $2^{3k}-1$.
$⟹2^{3k}-1=7M$, where $M\in N$.
$⟹2^{3k}=7M+1\to \left(1\right)$
Now consider $2^{3(k+1)}-1=2^{3k+3}-1=2^{3k}{.2}^3-1$
$=8(7M+1)-1$ using (1)
$=56M+7$ (As $2^{3k}=7M+1$)
$\therefore 2^{3(k+1)}-1=7(8M+1)$
$⟹$ 7 is a factor of $2^{3(k+1)}-1$
$⟹$ P(n) is true for $n=k+1$
$\therefore$ By the principle of mathematical induction, P(n) is true for all natural numbers n.
Hence, 7 is a factor of $2^{3n}-1$