3. The sequence of real numbers $\left(a_{n}\right)$ is defined by $a_{1}=5 ; a_{2}=19$ and for $n \geq 3$, $a_{n}=5 a_{n-1}-6 a_{n-2}$. Find $a_{2007}$.
Solution. Using the principle of mathematical induction, we will prove that

$$
a_{n}=3^{n+1}-2^{n+1}, n \geq 1
$$

Specifically, $a_{1}=3^{2}-2^{2} ; a_{2}=3^{3}-2^{3}$. Let for $k \leq n-1$ it holds that $a_{k}=3^{k+1}-2^{k+1}$. Then

$$
\begin{aligned}
a_{n} & =5 a_{n-1}-6 a_{n-2}=(3+2)\left(3^{n}-2^{n}\right)-3 \cdot 2\left(3^{n-1}-2^{n-1}\right) \\
& =3^{n+1}+2 \cdot 3^{n}-3 \cdot 2^{n}-2^{n+1}-2 \cdot 3^{n}+3 \cdot 2^{n}=3^{n+1}-2^{n+1}
\end{aligned}
$$

Therefore, it follows that $a_{2007}=3^{2008}-2^{2008}$.