Question

$(2^{3n}-1)$ will be divisible by $(\forall n\in N)$

• A. $25$
• B. $8$
• C. $7$
• D. $3$

Answer 1

The correct option is C $7$

$2^{3n}=(8{)}^n=(1+7{)}^n$
$={}^{1+}{C}_0+{}^n{C}_{1}7+{}^n{C}_2{7}^2+\cdots +{}^n{C}_n{7}^n$
$\Rightarrow 2^{3n}-1=7[{}^n{C}_1+{}^n{C}_{2}7+\cdots +{}^n{C}_n{7}^{n-1}]$
$\therefore$ divisible by $7$.