Solution: Recall that if $\vert z \vert \lt 1$, then \[ \sum_{n=0}^{\infty} z^n = \frac{1}{1-z}. \] We will use this for evaluating the given expression. Take a neighbourhood of $0$ such that $\vert z + z^2 \vert \lt 1$. Consider, \begin{align*} & \frac{1}{1 - (z + z^2)} = \sum_{n=0}^{\infty} \left( z + z^2 \right) ^n \\ \implies & \frac{z}{1 - z - z^2} = \sum_{n=0}^{\infty} z\left( z + z^2 \right) ^n. \end{align*} According to the given condition, \begin{align*} \sum_{n=0}^{\infty} z\left( z + z^2 \right) ^n = \sum_{n=0}^{\infty} a_n z^n. \end{align*}

Therefore, $a_5$ and $a_6$ will be the coefficient of $z^5$ and $z^6$ in $\sum_{n = 0} ^{\infty} z \left( z + z^2 \right) ^n$, respectively. Equivalently, \begin{align*} a_5 & = \text{ coefficient of $z^4$ in} \sum_{n=0}^{\infty} \left( z + z^2 \right) ^n \\ & = \text{ coefficient of $z^4$ in} \left[ \left( z + z^2 \right) ^2 + \left( z + z^2 \right) ^3 + \left( z + z^2 \right) ^4 \right] \\ & = \text{ coefficient of $z^0(z^2)^2$ in} \left( z + z^2 \right) ^2 \\ & \kern 0.5cm + \text{ coefficient of $z^2(z^2)^1$ in} \left( z + z^2 \right) ^3 \\ & \kern 0.5cm + \text{ coefficient of $z^4(z^2)^0$ in} \left( z + z^2 \right) ^4 \\ & = \binom{2}{2} + \binom{3}{1} + \binom{4}{0} \\ & = 1 + 3 + 1 = 5. \end{align*}

Similarly, \begin{align*} a_6 & = \text{ coefficient of $z^5$ in} \sum_{n=0}^{\infty} \left( z + z^2 \right) ^n \\ & = \text{ coefficient of $z^5$ in} \left[ \left( z + z^2 \right) ^3 + \left( z + z^2 \right) ^4 + \left( z + z^2 \right) ^5 \right] \\ & = \text{ coefficient of $z^1(z^2)^2$ in} \left( z + z^2 \right) ^3 \\ & \kern 0.5cm + \text{ coefficient of $z^3(z^2)^1$ in} \left( z + z^2 \right) ^4 \\ & \kern 0.5cm + \text{ coefficient of $z^5(z^2)^0$ in} \left( z + z^2 \right) ^5 \\ & = \binom{3}{2} + \binom{4}{3} + \binom{5}{0} \\ & = 3 + 4 + 1 = 8. \end{align*} Thus, \[ a_5 + a_6 = 5 + 8 = 13. \]