Proving the fractional equation: $\{2^{n-1}\sqrt{3}\}=0.b_nb_{n+1}\ldots_{(2)}$ 
Prove that $$\{2^{n-1}\sqrt{3}\}=0.b_nb_{n+1}\ldots_{(2)}$$ where  $\sqrt 3 = 1.b_1b_2b_3 \dots _{(2)}$.
  (Note: $\{x\} = x - \lfloor x \rfloor$ denotes the fractional part of $x$.)

I am not sure how to prove this result because of the fractional part. Should I expand $\sqrt{3}$ and then multiply by $2^{n-1}$?

From
$$
\sqrt 3 = 1.b_1b_2b_3 \cdots _{\,(2)} \tag1
$$ one gets, using $b_n \in \{0,1\}$,
$$
\begin{align}
\left\{2^{n-1}\sqrt 3\right\} &=\left\{2^{n-1}\left( 1+\frac{b_1}{2}+\frac{b_2}{2^2}+\cdots+\frac{b_{n-1}}{2^{n-1}}+\frac{b_n}{2^{n}}+\frac{b_{n+1}}{2^{n+1}}+\cdots \right)\right\}
\\\\&=\left\{2^{n-1}\left( 1+\frac{b_1}{2}+\frac{b_2}{2^2}+\cdots+\frac{b_{n-1}}{2^{n-1}}\right)+2^{n-1}\left(\frac{b_n}{2^{n}}+\frac{b_{n+1}}{2^{n+1}}+\cdots \right)\right\}
\\\\&=\left\{\underbrace{2^{n-1}+2^{n-2}b_1+2^{n-3}b_2+\cdots+b_{n-1}}_{\text{integer part}} +\overbrace{\frac{b_n}{2}+\frac{b_{n+1}}{2^2}+\cdots}^{\text{fractional part}}\right\}
\\\\&=\frac{b_n}{2}+\frac{b_{n+1}}{2^2}+\cdots
\\\\&=0.b_nb_{n+1}\cdots_{\,(2)}
\end{align}
$$ as announced.
