Question

The integer just greater than $(3+\sqrt{5})^{(2 n)}$ is divisible by, $(n \in N)$,

• A. $2^{n -1}$
• B. $2^{n + 1}$
• C. $2^{n + 2}$
• D. Not divisible by 2

Answer 1

Answer: (B)

$R=(3+\sqrt{5})^{2 n}, G=(3-\sqrt{5})^{2 n}$
Let $[R]+1=I$ ( [ ] greatest integer function).
$\Rightarrow R + G = I (\because 0< G < l )$
$\Rightarrow (3+\sqrt{5})^{2 n}+(3-\sqrt{5})^{2 n}=1$
Seeing the option put $n=1$
$\Rightarrow I =28$ is divisible by $4$ i.e $2^{ n +1}$