Let $\alpha$ and $\beta$ are roots of $x^2-6(t^2-2t+2)x-2=0$ with $\alpha >\beta .$ If $a_n={\alpha }^n-{\beta }^n$ for $n\ge 1,$ then find the minimum value of $\frac{a_{100}-2a_{98}}{a_{99}}$ (where $t\in R$ ) 

Let $t^2-2t+2=k\Rightarrow {\alpha }^2-6k\alpha -2=0$

${\alpha }^2-2=6k\alpha$

$a_{100-2a_{98}}={\alpha }^{100}-2g{a}^{98}-{\beta }^{100}+2{\beta }^{98}\Rightarrow {\alpha }^{98}({\alpha }^2-2)-{\beta }^{98}({\beta }^2-2)=6k({\alpha }^{99}-{\beta }^{99})$

$a_{100}-2a_{98}=6k.a_{99}$

$\frac{a_{100}-2a_{98}}{a_{99}}=6k=6(t^2-2t+2)=6({(t-1)}^2+1)$

$\therefore$minimum value of $\frac{a_{100}-2a_{98}}{a_{99}}$ is 6