1．已知$S_{n}$是数列$\left\{ a_{n} \right\}$的前$n$项和，且$a_{1} = a_{2} = 1$，$a_{n} = 2a_{n - 1} + 3a_{n - 2}$（$n \geq 3$），则下列结论正确的是（ ）
A．数列$\left\{ a_{n} - a_{n + 1} \right\}$为等比数列 B．数列$\left\{ a_{n + 1} + 2a_{n} \right\}$为等比数列
C．$S_{40} = \frac{1}{4}\left( 3^{20} - 1 \right)$ D．$a_{n} = \frac{3^{n - 1} + ( - 1)^{n - 1}}{2}$
解 由题意得：$a_{3} = 2a_{2} + 3a_{1} = 5$，$a_{4} = 2a_{3} + 3a_{2} = 10 + 3 = 13$，
由于$a_{1} - a_{2} = 0$，故数列$\left\{ a_{n} - a_{n + 1} \right\}$不是等比数列，A错误；
则$a_{2} + 2a_{1} = 1 + 2 = 3$，$a_{3} + 2a_{2} = 5 + 2 = 7$，$a_{4} + 2a_{3} = 13 + 10 = 23$，
由于$\frac{7}{3} \neq \frac{23}{7}$，故数列$\left\{ a_{n + 1} + 2a_{n} \right\}$不为等比数列，B错误；
$n \geq 3$时，$a_{n} = 2a_{n - 1} + 3a_{n - 2}$，即$a_{n} + a_{n - 1} = 3\left( a_{n - 1} + a_{n - 2} \right)$，
又$a_{1} + a_{2} = 1 + 1 = 2$，
故$\left\{ a_{n + 1} + a_{n} \right\}$为等比数列，首项为2，公比为3，
故$a_{n + 1} + a_{n} = 2 \times 3^{n - 1}$，
故$a_{2} + a_{1} = 2$，$a_{4} + a_{3} = 2 \times 3^{2}$，......，$a_{40} + a_{39} = 2 \times 3^{38}$，
以上20个式子相加得：$S_{40} = 2 \times \left( 1 + 3^{2} + 3^{4} + \cdots + 3^{38} \right) = 2 \times \frac{1 - 3^{40}}{1 - 9} = \frac{3^{40} - 1}{4}$，C错误；
因为$a_{n + 1} + a_{n} = 2 \times 3^{n - 1}$，所以$a_{n + 2} + a_{n + 1} = 2 \times 3^{n}$，两式相减得：
$a_{n + 2} - a_{n} = 2 \times 3^{n} - 2 \times 3^{n - 1} = 4 \times 3^{n - 1}$，
当$n = 2k$时，$a_{2k} - a_{2k - 2} = 4 \times 3^{2k - 3}$，$a_{2k - 2} - a_{2k - 4} = 4 \times 3^{2k - 5}$，......，$a_{4} - a_{2} = 4 \times 3$，
以上式子相加得：$a_{2k} - a_{2} = 4 \times \left( 3 + 3^{3} + \cdots + 3^{2k - 3} \right) = 4 \times \frac{3 - 3^{2k - 1}}{1 - 9} = \frac{3^{2k - 1} - 3}{2}$，
故$a_{2k} = \frac{3^{2k - 1} - 3}{2} + a_{2} = \frac{3^{2k - 1} - 1}{2}$，而$a_{2} = 1$也符和该式，故$a_{2k} = \frac{3^{2k - 1} - 1}{2}$，
令$2k = n$得：$a_{n} = \frac{3^{n - 1} - 1}{2} = \frac{3^{n - 1} + ( - 1)^{n - 1}}{2}$，
当$n = 2k - 1$时，$a_{2k - 1} - a_{2k - 3} = 4 \times 3^{2k - 4}$，$a_{2k - 3} - a_{2k - 5} = 4 \times 3^{2k - 6}$，......，$a_{3} - a_{1} = 4 \times 3^{0}$，
以上式子相加得：$a_{2k - 1} - a_{1} = 4 \times \left( 3^{2k - 4} + 3^{2k - 6} + \cdots + 3^{0} \right) = 4 \times \frac{1 - 3^{2k - 2}}{1 - 9} = \frac{3^{2k - 2} - 1}{2}$，
故$a_{2k - 1} = \frac{3^{2k - 2} - 1}{2} + a_{1} = \frac{3^{2k - 2} + 1}{2}$，而$a_{1} = 1$也符号该式，故$a_{2k - 1} = \frac{3^{2k - 2} + 1}{2}$，
令$2k - 1 = n$得：$a_{n} = \frac{3^{n - 1} + ( - 1)^{n - 1}}{2}$，
综上：$a_{n} = \frac{3^{n - 1} + ( - 1)^{n - 1}}{2}$，D正确.故选：D
2．已知数列$\left\{ a_{n} \right\}$，$\left\{ b_{n} \right\}$满足$a_{1} = 2$，$b_{1} = \frac{1}{2}$，$\left\{ \begin{array}{r} a_{n + 1} = b_{n} + \frac{1}{a_{n}} \\ b_{n + 1} = a_{n} + \frac{1}{b_{n}} \end{array} \right.$，$n \in \text{N}^{*}$，则下列选项错误的是（ ）
A．$\frac{a_{50}}{b_{50}} = \frac{1}{4}$ B．$a_{50}b_{50} < 112$
C．$a_{50} + b_{50} = \frac{5}{2}\sqrt{a_{50}b_{50}}$ D．$\left| a_{50} - b_{50} \right| \leq 15$
解 选项A：$a_{2} = b_{1} + \frac{1}{a_{1}} = \frac{1}{2} + \frac{1}{2} = 1$，$b_{2} = a_{1} + \frac{1}{b_{1}} = 2 + 2 = 4$，则$\frac{a_{2}}{b_{2}} = \frac{1}{4}$
又$\frac{a_{n + 1}}{b_{n + 1}} = \frac{b_{n} + \frac{1}{a_{n}}}{a_{n} + \frac{1}{b_{n}}} = \frac{b_{n}}{a_{n}}$，则$\frac{a_{n + 1}}{b_{n + 1}} = \frac{b_{n}}{a_{n}} = \frac{a_{n - 1}}{b_{n - 1}}$，
则$\frac{a_{50}}{b_{50}} = \frac{a_{48}}{b_{48}} = \cdots = \frac{a_{2}}{b_{2}} = \frac{1}{4}$.判断正确；
选项B： $a_{n + 1}b_{n + 1} = (b_{n} + \frac{1}{a_{n}})(a_{n} + \frac{1}{b_{n}}) = a_{n}b_{n} + 2 + \frac{1}{a_{n}b_{n}} \geq 4$
（当且仅当$a_{n}b_{n} = 1$时等号成立）
则$a_{50}b_{50} = a_{1}b_{1} + 2 \times 49 + \frac{1}{a_{1}b_{1}} + \frac{1}{a_{2}b_{2}} + \cdots + \frac{1}{a_{49}b_{49}}$
$< 2 + 2 \times 49 + \frac{1}{4} \times 48 = 112$.判断正确；
选项C：$a_{n + 1} + b_{n + 1} = a_{n} + b_{n} + \frac{1}{a_{n}} + \frac{1}{b_{n}} = \left( a_{n} + b_{n} \right)\left( 1 + \frac{1}{a_{n}b_{n}} \right)$，
则$\left( a_{n + 1} + b_{n + 1} \right)^{2} = \left( a_{n} + b_{n} \right)^{2}\left( 1 + \frac{1}{a_{n}b_{n}} \right)^{2}$
则$\frac{\left( a_{n + 1} + b_{n + 1} \right)^{2}}{a_{n + 1}b_{n + 1}} = \frac{\left( a_{n} + b_{n} \right)^{2}}{a_{n}b_{n}} = \cdots = \frac{\left( a_{1} + b_{1} \right)^{2}}{a_{1}b_{1}} = \frac{25}{4}$，
则$a_{50} + b_{50} = \frac{5}{2}\sqrt{a_{50}b_{50}}$.判断正确；
选项D：$a_{n + 1} - b_{n + 1} = b_{n} - a_{n} + \frac{1}{a_{n}} - \frac{1}{b_{n}} = \left( b_{n} - a_{n} \right)\left( 1 + \frac{1}{a_{n}b_{n}} \right)$，
则$\left( a_{n + 1} - b_{n + 1} \right)^{2} = \left( b_{n} - a_{n} \right)^{2}\left( 1 + \frac{1}{a_{n}b_{n}} \right)^{2}$
则$\frac{\left( a_{n + 1} - b_{n + 1} \right)^{2}}{a_{n + 1}b_{n + 1}} = \frac{\left( a_{n} - b_{n} \right)^{2}}{a_{n}b_{n}} = \cdots = \frac{\left( a_{1} - b_{1} \right)^{2}}{a_{1}b_{1}} = \frac{9}{4}$，
而$a_{50}b_{50} = a_{1}b_{1} + 2 \times 49 + \frac{1}{a_{1}b_{1}} + \frac{1}{a_{2}b_{2}} + \cdots + \frac{1}{a_{49}b_{49}}$ $> 2 + 2 \times 49 = 100$，
则$\left| a_{50} - b_{50} \right| = \frac{3}{2}\sqrt{a_{50}b_{50}} > 15$.判断错误.故选：D