Question

If $P\left(n\right):{}^{\prime \prime }{2}^{2n}-1$ is divisible by $k$ for all $n\in N^{\prime \prime }$ is true, then the value of ${}^{\prime }{k}^{\prime }$ is

• A. 6
• B. 3
• C. 7
• D. 2

Answer 1

The correct option is B 3

$2^{2n}-1=4^n-1$
$=(1+3{)}^n-1$
$=(1+(\frac{n}{1})\cdot 3+(\frac{n}{2})\cdot 3^2+\cdots +3^n)-1$
$=(\frac{n}{1})\cdot 3+(\frac{n}{1})\cdot 3^2+\cdots +3^n$
$=3m,\text{for some}m\in N$