# $\alpha, \beta$ 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 $\frac{a_{100}-2a_{98}}{a_{99}}$

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.

• Are you sure the one of the $a_{99}$'s isn't supposed to be $a_{98}$ instead? – JimmyK4542 Dec 31 '16 at 5:05
• Is $t$ a constant or the independent variable that $x$ depends on? – infinitylord Dec 31 '16 at 5:06
• @JimmyK4542 thank you, made the correction – Osheen Sachdev Dec 31 '16 at 5:07
• @infinitylord a variable that $x$ depends on (but the answer would be independent of $t$) – Osheen Sachdev Dec 31 '16 at 5:09

$$x^2-6(t^2+2t+2)x-2 = 0$$
$$x^{100}-6(t^2+2t+2)x^{99}-2x^{98} = 0$$
$$x^{100}-2x^{98} = 6(t^2+2t+2)x^{99}$$
Since $\alpha$ and $\beta$ are roots of the quadratic, we have $$\alpha^{100}-2\alpha^{98} = 6(t^2+2t+2)\alpha^{99}$$ and $$\beta^{100}-2\beta^{98} = 6(t^2+2t+2)\beta^{99}.$$