A little twist

Let's find an equation with roots $c=\alpha^2$ and $d=\beta^2$ , to do this let $y=x^2$ , so $x=\sqrt{y}$ :

$2y-5\sqrt{y}+1=0 \\ 2y+1=5\sqrt{y} \\ 4y^2+4y+1=25y \\ 4y^2-21y+1=0$

Now, $S_n=(c)^n+(d)^n$ , and since they are roots:

$4c^2-21c+1=0 \\ 4d^2-21d+1=0$

We can multiply both sides of each equation by $c^{2015}$ and $d^{2015}$ :

$4c^{2017}-21c^{2016}+c^{2015}=0 \\ 4d^{2017}-21d^{2016}+d^{2015}=0$

Finally, add them:

$4(c^{2017}+d^{2017})-21(c^{2016}+d^{2016})+c^{2015}+d^{2015}=0 \\ 4S_{2017}-21S_{2016}+S_{2015}=0 \\ 4S_{2017}+S_{2015}=21S_{2016} \\ \dfrac{4S_{2017}+S_{2015}}{S_{2016}}=\boxed{21}$

The required quantity is $\frac{4\alpha^{4034}+4\beta^{4034}+\alpha^{4030}+\beta^{4030}}{\alpha^{4032}+\beta^{4032}}$ . In the numerator, we can pull out common factors in both alpha and beta and it becomes $\frac{\alpha^{4030}(4\alpha^{4}+1)+\beta^{4030}(4\beta^{4}+1)}{\alpha^{4032}+\beta^{4032}}$ .

Now given alpha and beta satisfy that quadratic equation, we have $2\alpha^2-5\alpha+1=0$ . Taking $5\alpha$ to the RHS and squaring and simplifying the expression, we get $4\alpha^4+1=21\alpha^2$ and a similar expression for beta. Substitute this in the "factored" expression and note that the alpha and beta powers will cancel each other leaving behind 21

From Vieta's formulas, we have: $\begin{cases} \alpha + \beta = \frac{5}{2} \\ \alpha \beta = \frac{1}{2} \end{cases}$

Using Newton's sums method, we have:

$\begin{aligned} \alpha^{4032} + \beta^{4032} & = (\alpha + \beta) (\alpha^{4031} + \beta^{4031}) - \alpha \beta (\alpha^{4030} + \beta^{4030}) \\ \Rightarrow S_{2016} & = \frac{5}{2} (\alpha^{4031} + \beta^{4031}) - \frac{1}{2} S_{2015} \end{aligned}$

$\color{#3D99F6}{\Rightarrow 5(\alpha^{4031} + \beta^{4031}) = 2S_{2016} + S_{2015}}$

$\begin{aligned} \alpha^{4034} + \beta^{4034} & = (\alpha + \beta) (\alpha^{4033} + \beta^{4033}) - \alpha \beta (\alpha^{4032} + \beta^{4032}) \\ \Rightarrow S_{2017} & = \frac{5}{2} (\alpha^{4033} + \beta^{4033}) - \frac{1}{2} S_{2016} \\ 2 S_{2017} & = 5 \left(\frac{5}{2} S_{2016} - \frac{1}{2}(\alpha^{4031} + \beta^{4031})\right) - S_{2016} \\ 4 S_{2017} & = 25 S_{2016} - \color{#3D99F6}{5(\alpha^{4031} + \beta^{4031})} - S_{2016} \\ 4 S_{2017} & = 25 S_{2016} - \color{#3D99F6}{2S_{2016} - S_{2015}} - S_{2016} \\ 4 S_{2017} & = 21 S_{2016} - S_{2015} \\ \Rightarrow \frac{4 S_{2017}+S_{2015}}{S_{2016}} & = \boxed{21} \end{aligned}$

4S(n)=21S (n-1)-S (n-2)....prove it by breaking a^k+b^k=(a^(k-1)+b^(k-1))(a+b)-ab (a^(k-2)+b^(k-2))... Then the problem is trivial.