Question

Let $R=(5 \sqrt{5}+11)^{2 n+1}$ and $f=R-[R]$ where [.] denotes the greatest integer function then value of $R f$ is

• A. $2^{2 n+1}$
• B. $4^{2 n+1}$
• C. $2^{4 n+1}$
• D. $4^{2 n}$

Answer 1

Answer: (B)

$R=(5 \sqrt{5}+11)^{2 n+1}=1+f, 0 < f < 1$
Let $f^{\prime}=(5 \sqrt{5}-11)^{2 n+1}, 0 < f^{\prime} < 1$
$\therefore(5 \sqrt{5})^{2 n+1}+{ }^{2 n+1} C_1(5 \sqrt{5})^{2 n} .11+\ldots=I+f$
$\because-1 < f-f^{\prime} < 1$
$\therefore(5 \sqrt{5})^{2 n+1}-{ }^{2 n+1} C_1(5 \sqrt{5})^{2 n} \cdot 11+\ldots . f^{\prime}$
$\therefore 2\left({ }^{2 n+1} C_1 \cdot(\sqrt{5} 5)^{2 n} \cdot 11+\ldots.\right)=I+f-f^{\prime}=I$
$\therefore R f=R f^{\prime}=(5 \sqrt{5}+11)^{2 n+1} \cdot(5 \sqrt{5}-11)^{2 n+1}$
$=(125-121)^{2 n+1}=4^{2 n+1}$