Question

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

• A. $4^{2n+1}$
• B. $2^{2n+1}$
• C. $2^n$
• D. $8^{n+1}$

Answer 1

The correct option is A $4^{2n+1}$

Here $f=R-\left[R\right]$ is the fractional part of $R$ thus if $I$ is the integral part of $R$ Then
$R=I+f=(5\sqrt{5}+11{)}^{2n+1}and0<f<1Let{f}^{\prime }=(5\sqrt{5}-11{)}^{2n+1}.Then0<f^{\prime }<1(as5\sqrt{5}-11<1)$
Now $I+f-f^{\prime }=(5\sqrt{5}+11{)}^{2n+1}-(5\sqrt{5}-11{)}^{2n+1}=2{[}^{2n+1}{C}_1(5\sqrt{5}{)}^{2n}\times 11{+}^{2n+1}{C}_3(5\sqrt{5}{)}^{2n-2}\times {11}^3+\cdots ]$
= an even integer
$\Rightarrow f-f^{\prime }$ must also be an integer
$\Rightarrow f-f^{\prime }=0,(\because 0<f<1,0<f^{\prime }<1,\therefore -1<f-f^{\prime }<1)$
$\Rightarrow f=f^{\prime }$
$\therefore RF=R{F}^{\prime }=(5\sqrt{5}+11{)}^{2n+1}(5\sqrt{5}-11{)}^{2n+1}=(125-121{)}^{2n+1}=4^{2n+1}$