Recurrence problem for $a_5$ Assume that the sequence $\{a_0,a_1,a_2,\ldots\}$ satisﬁes the recurrence $a_{n+1} = a_n + 2a_{n−1}$. We know that $a_0 = 4$ and $a_2 = 13$. What is $a_5$?

From the recurrence you get $a_1=5$. Write the recurrence as:
$$
a_{n + 2} = a_{n + 1} + 2 a_n \qquad a_0 = 4, a_1 = 5
$$
Define the generating function:
$$
A(z) = \sum_{n \ge 0} a_n z^n
$$
Multiply the recurrence by $z^n$ and sum over $n \ge 0$. Notice that:
\begin{align}
\sum_{n \ge 0} a_{n + 1} z^n &= \frac{A(z) - a_0}{z} \\
\sum_{n \ge 0} a_{n + 2} z^n &= \frac{A(z) - a_0 - a_1 z}{z^2} \\
\end{align}
to get:
$$
\frac{A(z) - 4 - 5 z}{z^2} = \frac{A(z) - 4}{z} + 2 A(z)
$$
Solving this equation for $A(z)$, and splitting into partial fractions:
$$
A(z) = \frac{4 + z}{1 - z - 2 z^2} = \frac{1}{1 + z} + \frac{3}{1 - 2 z}
$$
This are just geometric series:
$$
a_n = (-1)^n + 3 \cdot 2^n
$$
So the value sought is $a_5 = (-1)^5 + 3 \cdot 2^5 = 95$
