Question

If $(6\sqrt{6}+14{)}^{2n+1}=N$ and $F=N-\left[N\right]$; where $\left[N\right]$ denotes greatest integer $\le N$, then $NF$ is equal to

• A. ${20}^{2n+1}$
• B. ${10}^{2n+1}$
• C. ${20}^{2n}$
• D. ${40}^{2n+1}$

Answer 1

The correct option is A ${20}^{2n+1}$

Let

$(6\sqrt{6}+14{)}^{2n+1}=N=I+F$

where I is a positive integer and $0<F<1$

clearly $(6\sqrt{6}-14{)}^{2n+1}=F^{\prime }$ where $0<F^{\prime }<1$

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

$(x+a{)}^n={}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}a+\cdots$

$(x-a{)}^n={}^n{C}_0{x}^n-{}^n{C}_1{x}^{n-1}a+{}^n{C}_2{x}^{n-2}{a}^2-\cdots$

$(x+a{)}^n-(x-a{)}^n=2({}^n{C}_1{x}^{n-1}a+{}^n{C}_3{x}^{n-3}{a}^3+\dots )$

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

$=2\left[{}^{2n+1}{C}_1\right(6\sqrt{6}{)}^{2n}\times 14+{}^{2n+1}{C}_3(6\sqrt{6}{)}^{2n-2}\times (14{)}^3$

$I+F-F^{\prime }$ is an even integer

$F-F^{\prime }=0$

$\Rightarrow F=F^{\prime },0<F<1$ and $0<F^{\prime }<1$

I is an even integer

$RF=R{F}^{\prime }$ or $NF=N{F}^{\prime }=(6\sqrt{6}+14{)}^{2n+1}(6\sqrt{6}-14{)}^{2n+1}$

$={\left[\right(6\sqrt{6}{)}^2-\left(14{)}^2\right]}^{2n+1}$

$=(216-196{)}^{2n+1}=(20{)}^{2n+1}$

Hence, (A) is the correct option.