Question

If $x=(\sqrt{3}+1{)}^n,$ where $n$ is odd positive integer, then $\left[x\right]$ is
(where $\left[x\right]$ denotes the greatest integer less than or equal to $x$)

• A. $2k$ where $k\in I$
• B. $2k+1$ where $k\in I$
• C. $4^n$
• D. $8^n$

Answer 1

The correct option is A $2k$ where $k\in I$

$x=(\sqrt{3}+1{)}^n=[x]+f$
$\therefore 0<f<1$
Let $f^{\prime }=(\sqrt{3}-1{)}^n$
$\Rightarrow 0<f^{\prime }<1\therefore -1<f-f^{\prime }<1$

$\left[x\right]+f-f^{\prime }=(\sqrt{3}+1{)}^n-(\sqrt{3}-1{)}^n=(1+\sqrt{3}{)}^n+(1-\sqrt{3}{)}^n(\because n\text{odd positive integer})=2[1+{}^n{C}_2(\sqrt{3}{)}^2+{}^n{C}_4(\sqrt{3}{)}^4+\cdots ]=2k,k\in I\Rightarrow f-f^{\prime }=2k-\left[x\right]\Rightarrow f-f^{\prime }=\text{integer}$

As $-1<f-f^{\prime }<1$, so
$f-f^{\prime }=0$
So,
$\left[x\right]=2k,k\in I$