Solution:

Answer: $\frac{13}{3}$ By using standard methods for solving linear recurrences $\{ \}^2$, we see that this recurrence has a characteristic polynomial of $x^{2}-\frac{5}{2} x+1=\left(x-\frac{1}{2}\right)(x-2)$, hence $a_{n}(x)=c(x) \cdot 2^{n}+d(x) \cdot 2^{-n}$ for some polynomials $c$ and $d$. Plugging in $n=1$ and $n=2$ gives
$$2 c(x)+\frac{1}{2} d(x)=3 x^{2}-x$$
and
$$4 c(x)+\frac{1}{4} d(x)=3 x^{2}-7 x+3$$
Subtracting the first equation from two times the second equation gives $6 c(x)=3 x^{2}-13 x+6$, so $c(x)=\frac{3 x^{2}-13 x+6}{6}$. As $n$ grows large, the $c(x) 2^{n}$ term dominates compared to the $d(x) 2^{-n}$ term, so the roots of $a_{n}(x)$ converge to the roots of $c(x)$. Thus the roots of $a_{n}(x)$ converge to the roots of $3 x^{2}-13 x+6$, which by Vieta's formula $\{ \}^3$ have a sum of $\frac{13}{3}$.