已知数列 $\left\{ {{a_n}} \right\}$、$\left\{ {{b_n}} \right\}$ 满足 ${a_{n + 1}} = - {a_n} - 2{b_n}$，且 ${b_{n + 1}} = 6{a_n} + 6{b_n}$，又有 ${a_1} = 2$，${b_1} = 4$，求：
【难度】
【出处】
2004年复旦大学保送生招生测试
【标注】
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${a_n},{b_n}$；标注答案${a_n} = {2^{n + 3}} - 14 \cdot {3^{n - 1}}$，${b_n} = - 3 \cdot {2^{n + 2}} + 28 \cdot {3^{n - 1}}$解析设$$\alpha {b_{n + 1}} - {a_{n + 1}} = \beta \left( {\alpha {b_n} - {a_n}} \right),$$则$$\left( {6\alpha + 1} \right){a_n} + \left( {6\alpha + 2} \right){b_n} = - \beta {a_n} + \alpha \beta {b_n}.$$解方程$$\begin{cases}
6\alpha + 1 = - \beta ,\\
6\alpha + 2 = \alpha \beta ,\\
\end{cases}$$得$$(\alpha,\beta)=\left(-\dfrac 12,3\right),\left(-\dfrac 12,2\right),$$设辅助数列$${x_n} = - \dfrac{2}{3}{b_n} - {a_n},$$则$${x_1} = - \dfrac{{14}}{3},{x_n} = - \dfrac{{14}}{3} \cdot {3^{n - 1}};$$设辅助数列$${y_n} = - \dfrac{1}{2}{b_n} - {a_n},$$则$${y_1} = - 4,{y_n} = - 4 \cdot {2^{n - 1}}.$$于是解得\[\begin{split}{a_n} &= {2^{n + 3}} - 14 \cdot {3^{n - 1}},\\ {b_n} &= - 3 \cdot {2^{n + 2}} + 28 \cdot {3^{n - 1}}.\end{split}\] -
$\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_n}}}{{{b_n}}}$．标注答案$- \dfrac{1}{2}$解析结合第 $(1)$ 小题，可得\[\begin{split}\mathop {\lim }\limits_{n \to + \infty } \dfrac{{{a_n}}}{{{b_n}}}&= \mathop {\lim }\limits_{n \to + \infty } \dfrac{{16 \cdot {2^{n - 1}} - 14 \cdot {3^{n - 1}}}}{{ - 24 \cdot {2^{n - 1}} + 28 \cdot {3^{n - 1}}}} \\&= \mathop {\lim }\limits_{n \to + \infty } \dfrac{{16 \cdot {{\left( {\dfrac{2}{3}} \right)}^{n - 1}} - 14}}{{ - 24 \cdot {{\left( {\dfrac{2}{3}} \right)}^{n - 1}} + 28}} \\&= - \dfrac{1}{2}.\end{split}\]
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