Question

For each $n\in N,2^{3n}-1$ is divisible by

• A. $7$
• B. $8$
• C. $6$
• D. $16$
  where $N$ is a set of natural numbers

Answer 1

Answer: (A)

$7$

$2^{3n}-1={(1+7)}^n-1$
$=(1{+}^n{C}_{1}7{+}^n{C}_2{7}^2+\dots \infty )-1$
$=7({}^n{C}_1{+}^n{C}_{2}7+\dots )$
Hence, it is divisible by $7$.
Alternative
Let $P\left(n\right)=2^{3n}-1$
$P\left(1\right)=2^3-1=7$
$P\left(2\right)=2^6-1=63$
$P\left(3\right)=2^9-1=511$
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Hence, it is divisible by $7$.