$\alpha, \beta$ are the roots of $x^2-6(t^2-2t+2)x-2=0$ and $a_n=\alpha^n-\beta^n$ for $n\geq 1$, then find $\frac{a_{100}-2a_{98}}{a_{99}}$

Question

Let $\alpha, \beta$ be the roots of $x^2-6(t^2-2t+2)x-2=0$, with $\alpha>\beta$. If $a_n=\alpha^n-\beta^n$ for $n\geq 1$, then find the value of $\dfrac{a_{100}-2a_{98}}{a_{99}}$

Okay I have no idea as to how I can solve this question. I just need a small hint so that I can get a direction in order to proceed.

Answer 1

Hint: We can manipulate the quadratic equation as follows:

$$x^2-6(t^2+2t+2)x-2 = 0$$

$$x^{100}-6(t^2+2t+2)x^{99}-2x^{98} = 0$$

$$x^{100}-2x^{98} = 6(t^2+2t+2)x^{99}$$

Since $\alpha$ and $\beta$ are roots of the quadratic, we have $$\alpha^{100}-2\alpha^{98} = 6(t^2+2t+2)\alpha^{99}$$ and $$\beta^{100}-2\beta^{98} = 6(t^2+2t+2)\beta^{99}.$$