Question

If $\left[x\right]$ denotes the greatest integer less than or equal to $x$ and $F=R-\left[R\right]$ where $R=(5\sqrt{5}+11{)}^{2n+1}$ then $RF$ is equal to :

• A. $4^{2n+1}$
• B. $4^{2n}$
• C. $4^{2n-1}$
• D. None of these

Answer 1

Answer: (A)

$4^{2n+1}$

Since $(5\sqrt{5}+11)(5\sqrt{5}-11)=4$
$\Rightarrow 5\sqrt{5}-11=\frac{4}{5\sqrt{5}+11}$
Now $0<5\sqrt{5}-11<1$
$\Rightarrow 0<(5\sqrt{5}+11{)}^{2n+1}<1$ for $nϵI^{+}$
Again $(5\sqrt{5}+11{)}^{2n+1}-(5\sqrt{5}+11{)}^{2n+1}$
$=2{\{}^{2n+1}{C}_1(5\sqrt{5}{)}^{2n}\cdot 11{+}^{2n+1}{C}_3(5\sqrt{5}{)}^{2n-2}\cdot {11}^3+..{.}^{2n+1}{C}_{2n+1}{11}^{2n+1}\}$
$=2k$ (for some $kϵI^{+})$
let $F^{\prime }=(5\sqrt{5}-11{)}^{2n+1}$ then $\left[R\right]+F-F^{\prime }=2k$
$\Rightarrow F-F^{\prime }=2k-\left[R\right]\Rightarrow F-F^{\prime }ϵI$
But $0\le F<1,0<F^{\prime }<1\Rightarrow -1<F-F^{\prime }<1$
$\therefore RF=R{F}^{\prime }=(5\sqrt{5}+11{)}^{2n+1}$
$=\left[\right(5\sqrt{5}{)}^2-{11}^2{]}^{2n+1}=4^{2n+1}$