回復 1# bch0722b 的帖子
\(\begin{align}
& \left\{ \begin{align}
& {{a}_{n+1}}=2{{a}_{n}}+3{{b}_{n}}\cdots \left( 1 \right) \\
& {{b}_{n+1}}={{a}_{n}}+2{{b}_{n}}\cdots \left( 2 \right) \\
\end{align} \right. \\
& \left( 1 \right)\times 2-\left( 2 \right)\times 3 \\
& 2{{a}_{n+1}}-3{{b}_{n+1}}={{a}_{n}} \\
& 2{{a}_{n+1}}-\left( {{a}_{n+2}}-2{{a}_{n+1}} \right)={{a}_{n}} \\
& {{a}_{n+2}}=4{{a}_{n+1}}-{{a}_{n}} \\
\end{align}\)
而\({{a}_{1}}=2,{{a}_{2}}=7\)
同理也可得
\({{b}_{n+2}}=4{{b}_{n+1}}-{{b}_{n}}\)，很巧，跟上面一樣
而\({{b}_{1}}=1,{{b}_{2}}=4\)
……
