If \[a=1+\left( \sqrt{3}-1 \right)+{{\left( \sqrt{3}-1 \right)}^{2}}+{{\left( \sqrt{3}-1 \right)}^{3}}+...\] and \[ab=1\], then $a$ and $b$are the roots of the equation
A. \[{{x}^{2}}+4x-1=0\]
B. \[{{x}^{2}}-4x-1=0\]
C. \[{{x}^{2}}+4x+1=0\]
D. \[{{x}^{2}}-4x+1=0\]

Hint: In this question, we have to find the equation that is formed by the roots $a$ and $b$ where the root $a$ is the sum of the given series. The root $b$ is obtained by the given condition. Since the given series is a geometric series, we can find the sum of the infinite G.P for $a$ value.

Formula Used: If the series is a geometric series, then the sum of the $n$ terms is calculated by
 ${{S}_{n}}=\dfrac{a({{r}^{n}}-1)}{r-1}$ where $r=\dfrac{{{a}_{n}}}{{{a}_{n-1}}}$
Here ${{S}_{n}}$ - Sum of the $n$ terms of the series; $n$ - Number of terms; $a$ - First term in the series; $r$ - Common ratio.
The sum of the infinite G.P is calculated by
 ${{S}_{\infty }}=\dfrac{a}{1-r}$
Here ${{S}_{\infty }}$ - Sum of the infinite terms of the series
The quadratic equation formed by two roots $a$ and $b$ is
$(x-a)(x-b)=0$

Complete step by step solution: Given series is
\[a=1+\left( \sqrt{3}-1 \right)+{{\left( \sqrt{3}-1 \right)}^{2}}+{{\left( \sqrt{3}-1 \right)}^{3}}+...\]
This is a geometric series. So, the sum of the $n$ terms is
\[\begin{align}
  & a=\dfrac{1}{1-\left( \sqrt{3}-1 \right)} \\
 & \Rightarrow a=\dfrac{1}{2-\sqrt{3}} \\
\end{align}\]
To simplify this complex number $2-\sqrt{3}$, multiplying and dividing this by its conjugate $2+\sqrt{3}$.
Then, we get
\[\begin{align}
  & a=\dfrac{1}{2-\sqrt{3}}\times \dfrac{2+\sqrt{3}}{2+\sqrt{3}} \\
 & \text{ }=\dfrac{2+\sqrt{3}}{4-3}=2+\sqrt{3} \\
\end{align}\]
But we have \[ab=1\]. From this, we get
 \[\begin{align}
  & ab=1 \\
 & \Rightarrow b=\dfrac{1}{a} \\
\end{align}\]
On substituting $a$ value, we get
\[\begin{align}
  & b=\dfrac{1}{a} \\
 & \text{ }=\dfrac{1}{2+\sqrt{3}} \\
\end{align}\]
Since the obtained value is a complex number, we need to multiply and divide this by its complex conjugate.
I.e.,
\[\begin{align}
  & b=\dfrac{1}{2+\sqrt{3}}\times \dfrac{2-\sqrt{3}}{2-\sqrt{3}} \\
 & \text{ }=\dfrac{2-\sqrt{3}}{4-3}=2-\sqrt{3} \\
\end{align}\]
Since we have the roots $a$ and $b$, the required equation is
\[\begin{align}
  & (x-a)(x-b)=0 \\
 & \Rightarrow (x-(2+\sqrt{3}))(x-(2-\sqrt{3}))=0 \\
 & \Rightarrow \left( {{(x-2)}^{2}}-{{(\sqrt{3})}^{2}} \right)=0 \\
 & \Rightarrow {{x}^{2}}-4x+4-3=0 \\
 & \Rightarrow {{x}^{2}}-4x+1=0 \\
\end{align}\]

Option ‘D’ is correct

Note: Here we need to use the complex conjugates for simplifying the obtained complex numbers. So, that we can easily frame the required equation.