Question

For all natural numbers n, $2^{3n}-7n-1$ is divisible by

• A. 64
• B. 36
• C. 49
• D. 25

Answer 1

The correct option is C

49

Substitute n=1 in $2^{3n}-7n-1$, we get
$2^3-7-1=0\to \text{divisible by all positive integers}$

Substitute n=2 in $2^{3n}-7n-1$, we get
$2^6-14-1=49$
$LetP\left(n\right):2^{3n}-7n-1\text{is divisible by 49}$
P(2) is true.
Assume P(k) is true
$2^{3k}-7k-1=49m$
Substituting k+1 in place of n, we get
$2^{3k+3}-7(k+1)-1={8.2}^{3k}-7k-8=8.(2^{3k}-7k-1)+7.7k=49(8m+k)\to \text{divisible by 49}$
P(k+1) is true

Hence, P(n) is true.