Question

Let $\alpha ,\beta ,\gamma$ be the roots of the cubic equation $x^3+a{x}^2+bx+c=0,$ which (taken in given order) are in G.P. If $\alpha$ and $\beta$ are such that $\left|\begin{array}{ccc}2& 1& 2\\ 1+\alpha & \alpha & \beta \\ 4-\beta & 3-\beta & \alpha +1\end{array}\right|=0,$ then the value of $\sum _{r=1}^{100}({\left(\frac{\alpha }{\beta }\right)}^r+{\left(\frac{a}{b}\right)}^r)$ is

• A. $\frac{1}{3}(1-\frac{1}{2^{101}})$
• B. $\frac{1}{3}(1-\frac{1}{2^{100}})$
• C. $\frac{2}{3}(1-\frac{1}{2^{100}})$
• D. $\frac{2}{3}(1-\frac{1}{2^{101}})$

Answer 1

The correct option is C $\frac{2}{3}(1-\frac{1}{2^{100}})$

$\Delta =\left|\begin{array}{ccc}2& 1& 2\\ 1+\alpha & \alpha & \beta \\ 4-\beta & 3-\beta & \alpha +1\end{array}\right|$

$C_1\to C_1-C_2$
$\Delta =\left|\begin{array}{ccc}1& 1& 2\\ 1& \alpha & \beta \\ 1& 3-\beta & \alpha +1\end{array}\right|$

$C_2\to C_2-C_1,C_3\to C_3-2C_1$
$\Delta =\left|\begin{array}{ccc}1& 0& 0\\ 1& \alpha -1& \beta -2\\ 1& 2-\beta & \alpha -1\end{array}\right|=(\alpha -1{)}^2+(\beta -2{)}^2$

$(\alpha -1{)}^2+(\beta -2{)}^2=0\Rightarrow \alpha =1,\beta =2$

As $\alpha ,\beta ,\gamma$ are in G.P., so
${\beta }^2=\alpha \gamma \Rightarrow \gamma =4$
Therefore, the equation is
$(x-1)(x-2)(x-4)=0\Rightarrow x^3-7x^2+14x-8=0$
$a=-7,b=14,c=-8$

Now, $\sum _{r=1}^{100}{\left(\frac{\alpha }{\beta }\right)}^r+{\left(\frac{a}{b}\right)}^r$
$=\sum _{r=1}^{100}{\left(\frac{1}{2}\right)}^r+{\left(\frac{-1}{2}\right)}^r=\sum _{r=1}^{50}2{\left(\frac{1}{2}\right)}^{2r}=2\times \frac{\frac{1}{4}(1-{\left(\frac{1}{4}\right)}^{50})}{(1-\frac{1}{4})}=\frac{2}{3}(1-\frac{1}{2^{100}})$