11. Let $\left\{a_{n}\right\}$ be a sequence such that $a_{1}=20, a_{2}=17$ and $a_{n}+2 a_{n-2}=3 a_{n-1}$. Determine the value of $a_{2017}-a_{2016}$.
Answer: $-3 \times 2^{2015}$
Solution: We have $a_{n}=3 a_{n-1}-2 a_{n-2} \Longrightarrow a_{n}-a_{n-1}=2\left(a_{n-1}-a_{n-2}\right)$. Repeated use of this recurrence relation gives $a_{2017}-a_{2016}=2^{2015}\left(a_{2}-a_{1}\right)=-3 \cdot 2^{2015}$.