Let $\alpha, \beta$ be the roots of the equation $a x^2+b x+c=0, a, b, c$ real and $s_n=\alpha^n+\beta^n$ and $\left|\begin{array}{ccc}3 & 1+s_1 & 1+s_2 \ 1+s_1 & 1+s_2 & 1+s_3 \ 1+s_2 & 1+s_3 & 1+s_4\end{array}\right|=k \frac{(a+b+c)^2}{a^4}$ then $k=$

Question

Let $\alpha, \beta$ be the roots of the equation $a x^2+b x+c=0, a, b, c$ real and $s_n=\alpha^n+\beta^n$ and $\left|\begin{array}{ccc}3 & 1+s_1 & 1+s_2 \ 1+s_1 & 1+s_2 & 1+s_3 \ 1+s_2 & 1+s_3 & 1+s_4\end{array}\right|=k \frac{(a+b+c)^2}{a^4}$ then $k=$, $b^2-4 a c$, $b^2+4 a c$, $b^2+2 a c$, $4 a c-b^2$

MARKS App · Accepted Answer

Hint: $\left|\begin{array}{ccc}3 & 1+s_1 & 1+s_2 \ 1+s_1 & 1+s_2 & 1+s_3 \ 1+s_2 & 1+s_3 & 1+s_4\end{array}\right|$ $=\left|\begin{array}{ccc} 1+1+1 & 1+\alpha+\beta & 1+\alpha^2+\beta^2 \ 1+\alpha+\beta & 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 \ 1+\alpha^2+\beta^2 & 1+\alpha^3+\beta^3 & 1+\alpha^4+\beta^4 \end{array}\right|$ $\begin{aligned} & =\left|\begin{array}{ccc}1 & 1 & 1 \ 1 & \alpha & \beta \ 1 & \alpha^2 & \beta^2\end{array}\right| \times\left|\begin{array}{ccc}1 & 1 & 1 \ 1 & \alpha & \alpha^2 \ 1 & \beta & \beta^2\end{array}\right| \ & =\{(1-\alpha)(\alpha-\beta)(1-\beta)\}^2 \ & =(1-(\alpha+\beta)+\alpha \beta)^2(\alpha-\beta)^2 \ & =\left(1+\frac{b}{a}+\frac{c}{a}\right)^2\left(\frac{b^2}{a^2}-\frac{4 c}{a}\right) \ & =\frac{(a+b+c)^2}{a^2} \times \frac{b^2-4 a c}{a^2}=\left(b^2-4 a c\right) \frac{(a+b+c)^2}{a^4} \ & \therefore k=b^2-4 a c\end{aligned}$

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