Question

$R={(14+6\sqrt{6})}^{2n+1}$, then integral part of $R$ is an even integer and $Rf={20}^{2n+1}$ where $f$ is the fractional part of $R$.

• A. True
• B. False

Answer 1

The correct option is A True

$R=(14+6\sqrt{6}{)}^{2n+1}$

$\text{Let}{R}^{\prime }=(-14+6\sqrt{6}{)}^{2n+1}$

${\text{R-R}}^{\prime }=(6\sqrt{6}+14{)}^{2n+1}-(6\sqrt{6}-14{)}^{2n+1}$

$=2\left(\text{Sum of odd terms}\right)=\text{even Integer}$

$\text{Also}R=I+f\text{and}R-R^{\prime }=I+f-R^{\prime }$

$\text{will also be an even Integer}$

$\text{So}f-R^{\prime }=0\text{since}0<f<1,0<R^{\prime }<1$

$\text{Now}Rf=R{R}^{\prime }=$

$(6\sqrt{6}+14{)}^{2n+1}(6\sqrt{6}-14{)}^{2n+1}={20}^{2n+1}$