Let $\alpha$ and $\beta$ be the roots of $x^2-6 x-2=0$ with $\alpha\beta$. If $a_n=\alpha^n-\beta^n$ for $n \geq 1$, then the value of $\frac{a_{10}-2 a_8}{2 a_9}$ is

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Let $\alpha$ and $\beta$ be the roots of $x^2-6 x-2=0$ with $\alpha>\beta$. If $a_n=\alpha^n-\beta^n$ for $n \geq 1$, then the value of $\frac{a_{10}-2 a_8}{2 a_9}$ is

Aicharya · Accepted Answer

$\frac{a_{10}-2 a_8}{2 a_9}=\frac{\left(\alpha^{10}-\beta^{10}ight)-2\left(\alpha^8-\beta^8ight)}{2\left(\alpha^9-\beta^9ight)}$ $=\frac{\alpha^8\left(\alpha^2-2ight)-\beta^8\left(\beta^2-2ight)}{2\left(\alpha^9-\beta^9ight)}$ $\left[\because \alphaight.$ is root of $x^2-6 x-2=0$ $\left.\alpha^2-2=6 \alphaight]$ Also, $\beta$ is root of $x^2-6 x-2=0$ $ \begin{aligned} & \Rightarrow \quad \beta^2-2=6 \beta \ & =\frac{\alpha^8(6 \alpha)-\beta^8(6 \beta)}{2\left(\alpha^9-\beta^9ight)}=\frac{6\left(\alpha^9-\beta^9ight)}{2\left(\alpha^9-\beta^9ight)}=3 \end{aligned} $

Let $\alpha$ and $\beta$ be the roots of $x^2-6 x-2=0$ with $\alpha>\beta$. If $a_n=\alpha^n-\beta^n$ for $n…

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