Question

Let $R={(5\sqrt{5}+11)}^{2n+1}$ and $f=R-\left[R\right]$, then prove that $Rf=4^{2n+1}$ .

Answer 1

Given $R={(5\sqrt{5}+11)}^{2n+1}$ where $f=R-\left[R\right]$

Let us assume a number $f_1={(5\sqrt{5}-11)}^{2n+1}$

$\therefore 5\sqrt{5}-11$ lies between $0$ and $1$

$\Rightarrow 0<5\sqrt{5}-11<1$

$2n+1$ is a positive integer.

$\therefore 0<f_1<1$

$\Rightarrow R-f_1={(5\sqrt{5}+11)}^{2n+1}-{(5\sqrt{5}-11)}^{2n+1}$

$2({{2n+1}_C}_1{\left(5\sqrt{5}\right)}^{2n}\left(11\right)+{{2n+1}_C}_3{\left(5\sqrt{5}\right)}^{2n-2}{11}^3+{{2n+1}_C}_5{\left(5\sqrt{5}\right)}^{2n-3}{11}^5+...+{{2n+1}_C}_{2n+1}{11}^{2n+1})$

$\therefore \left[R\right]+f-f_1=$even

$\Rightarrow f-f_1=$integer

$\Rightarrow 0<f<1$ and $0<f_1<1$

$\Rightarrow f-f_1\in I$

$\Rightarrow f-f_1=0$ $\because f_1$ cannot be negative or $f_1$ cannot be negative.

$\Rightarrow f=f_1$

Now $Rf=R{f}_1$ since $f=f_1$

where $R={(5\sqrt{5}+11)}^{2n+1}$ since $Rf=R{f}_1$ where $R={(5\sqrt{5}+11)}^{2n+1}$

$Rf=R{f}_1={(5\sqrt{5}+11)}^{2n+1}{(5\sqrt{5}-11)}^{2n+1}$

$={({\left(5\sqrt{5}\right)}^2-121)}^{2n+1}$ using $(a+b)(a+b)=a^2-b^2$

$={(125-121)}^{2n+1}=4^{2n+1}$

Hence proved.