Question

If $\alpha$, $\beta$ are roots of $a{x}^2+bx+c=0$ and $D=b^2-4ac$, and $S_n=1+{\alpha }^n+{\beta }^n$ then
$\left|\begin{array}{ccc}{S}_0& S_1& S_2\\ S_1& S_2& S_3\\ S_2& S_3& S_4\end{array}\right|$

• A. $-1$
• B. $\frac{D{(a+b+c)}^3}{a^4}$
• C. $\frac{{(b^2-4ac)}^2}{a^4}$
• D. $\frac{D{(a+b+c)}^2}{a^4}$

Answer 1

Answer: (D)

$\frac{D{(a+b+c)}^2}{a^4}$

$\left|\begin{array}{ccc}{S}_0& S_1& S_2\\ S_1& S_2& S_3\\ S_2& S_3& S_4\end{array}\right|=\left|\begin{array}{ccc}3& 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2\\ 1+\alpha +\beta & 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3\\ 1+{\alpha }^2+{\beta }^2& 1+{\alpha }^3+{\beta }^3& 1+{\alpha }^4+{\beta }^4\end{array}\right|=\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|={\left|\begin{array}{ccc}1& 1& 1\\ 1& \alpha & \beta \\ 1& {\alpha }^2& {\beta }^2\end{array}\right|}^2={(1-\alpha )}^2{(1-\beta )}^2{(\alpha -\beta )}^2={(1-\alpha -\beta +\alpha \beta )}^2({(\alpha +\beta )}^2-4\alpha \beta )={(1-(-\frac{b}{a})+\frac{c}{a})}^2({(-\frac{b}{a})}^2-4\frac{c}{a})=\frac{D{(a+b+c)}^2}{a^4}$