Question

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

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

Answer 1

Answer: (C)

$4^{2n+1}$

Here $f=R-\left[R\right]$ is the fraction part of $R$. Thus if $I$ is the integral part of $R$ then $R=I+F={(5\sqrt{5}+11)}^{2n+1}$

and $0<f<1$

Now since $5\sqrt{5}-11=0.18<1$

Thus if $f^{{}^{\prime }}={(5\sqrt{5}-12)}^{2n+1}$ then ${0<f}^{{}^{\prime }}<1$

$\Rightarrow I+f-f^{{}^{\prime }}={(5\sqrt{5}+11)}^{2n+1}-{(5\sqrt{5}-11)}^{2n+1}$

$=2[{}^{2n+1}{C}_1{\left(5\sqrt{5}\right)}^{2n}\times 11{+}^{2n+1}{C}_3{\left(5\sqrt{5}\right)}^{2n-2}\times {11}^3+...]$...... is an Even integer

$f-f^{{}^{\prime }}$ must also be an integer

$f-f^{{}^{\prime }}=0\because 0<f<1,0<f^{{}^{\prime }}<1$

Thus $f=f^{{}^{\prime }}$

So, $Rf={Rf}^{{}^{\prime }}={(5\sqrt{5}+11)}^{2n+1}{(5\sqrt{5}-11)}^{2n+1}={(125-121)}^{2n+1}={\left(4\right)}^{2n+1}$