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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.

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    $\begingroup$ Are you sure the one of the $a_{99}$'s isn't supposed to be $a_{98}$ instead? $\endgroup$ – JimmyK4542 Dec 31 '16 at 5:05
  • $\begingroup$ Is $t$ a constant or the independent variable that $x$ depends on? $\endgroup$ – infinitylord Dec 31 '16 at 5:06
  • $\begingroup$ @JimmyK4542 thank you, made the correction $\endgroup$ – Osheen Sachdev Dec 31 '16 at 5:07
  • $\begingroup$ @infinitylord a variable that $x$ depends on (but the answer would be independent of $t$) $\endgroup$ – Osheen Sachdev Dec 31 '16 at 5:09
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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 $$\beta^{100}-2\beta^{98} = 6(t^2+2t+2)\beta^{99}.$$

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