浙江 2020 · 数学 q20

题面

20．已知数列{a_n}，{b_n}，{c_n}中，$a_{1} = b_{1} = c_{1} = 1 , \ c_{n + 1} = a_{n + 1} - a_{n} , \ c_{n + 1} = \frac{{\mathrm{b}}_{\mathrm{n}}}{{\mathrm{b}}_{\mathrm{n} + 2}} \bullet_{\mathrm{c}_{\mathrm{n}}} \ ( {n \in} \mathrm{N}^{*} )$

（Ⅰ）若数列$\left\{b_{n} \right\}$为等比数列，且公比$q {>} 0$，且$b_{1} + b_{2} = 6 b_{3}$，求q与$a_{n}$的通项公式；

（Ⅱ）若数列$\left\{b_{n} \right\}$为等差数列，且公差$d {>} 0$，证明：$c_{1} + c_{2} + \cdots + c_{n} < 1 + \frac{1}{d} .$

（Ⅰ）若数列$\left\{b_{n} \right\}$为等比数列，且公比$q {>} 0$，且$b_{1} + b_{2} = 6 b_{3}$，求q与$a_{n}$的通项公式；

（Ⅱ）若数列$\left\{b_{n} \right\}$为等差数列，且公差d＞0，证明：$c_{1} + c_{2} + \cdots + c_{n} < 1 + \frac{1}{d} .$

解析

分析

本题第（Ⅰ）题先根据等比数列的通项公式将$b_{2}{=} q , \ b_{3}{=} q^{2}$代入$b_{1} + b_{2} = 6 b_{3}$计算出公比 q 的值，然后根据等比数列的定义化简$c_{n + 1} = \frac{b_{n}}{b_{n + 2}} \bullet_{c_{n}}$可得$c_{n + 1} = 4 c_{\mathrm{n}}$，则可发现数列$\{{c}_{\mathrm{n}} \}$是以 1 为首项，4 为公比的等比数列，从而可得数列$\{c_{\mathrm{{n}}} \}$的通项公式，然后将通项公式代入$c_{n + 1} = a_{n + 1} - a_{n}$，可得$a_{n + 1} - a_{n} = c_{n + 1} = .$4ⁿ，再根据此递推公式的特点运用累加法可计算出数列$\left\{a_{n} \right\}$的通项公式；

第（Ⅱ）题通过将已知关系式$c_{n + 1} = \frac{b_{n}}{b_{n + 2}} \bullet_{c_{n}}$不断进行转化可构造出数列$\left\{b_{n} b_{n + 1} c_{\mathrm{n}} \right\}$，且可得到数列$\left\{b_{n} b_{n + 1} c_{\mathrm{n}} \right\}$是一个常数列，且此常数为 1+d，从而可得$b_{n} b_{n + 1} c_{\mathrm{n}}{=} 1 + d ,$，再计算得到$c_{\mathrm{n}} = \frac{1 + d}{b_{\mathrm{n}} b_{\mathrm{r} + 1}}$，根据等差数列的特点进行转化进行裂项，在求和时相消，最后运用放缩法即可证明不等式成立

解答

（Ⅰ）解：由题意，$b_{2} = q , \ b_{3} = q^{2}$

$\because b_{1} + b_{2} = 6 b_{3}, \therefore 1 + q = 6 q^{2},$

整理，得$6 q^{2} - q - 1 = 0$

解得$q = - \frac{1}{3} ( \frac{}{5} \pm )$，或$q {=} \frac{1}{2} ,$

$\therefore c_{n + 1} = \frac{b_{n}}{b_{n + 2}} \cdot c_{n} = \frac{\frac{1}{b_{n + 2}}}{\frac{b_{n}}{b_{n}}} \cdot c_{n} = \frac{1}{q^{2}} \cdot c_{n} = \frac{1}{(\frac{1}{2})^{2}} \cdot c_{n} = 4 \cdot c_{n},$

∴数列$\{{c}_{\mathrm{n}} \}$是以1为首项，4为公比的等比数列，

$\therefore \mathrm{c}_{\mathrm{n}} = 1 \cdot 4^{n^{-} 1} = 4^{n^{-} 1}, n \in \mathrm{N}^{*}.$

$\therefore a_{n + 1} - a_{n} = c_{n + 1} = 4^{n},$

则$a_{1} = 1 ,$

$a_{2} - a_{1} = 4^{1},$

$a_{3} - a_{2} = 4^{2},$

$a_{n} - a_{n - 1} = 4^{n^{-} 1},$

各项相加，可得

$a_{n} = 1 + 4^{1} + 4^{2} + \dots + 4^{n - 1} = \frac{1 - 4^{n}}{1 - 4} = \frac{4^{n} - 1}{3}.$

（Ⅱ）证明：依题意，由$c_{n + 1} = \frac{b_{\mathrm{n}}}{b_{\mathrm{n + 2}}} \bullet_{\mathrm{c}_{\mathrm{n}}} ( n \in \mathrm{N}^{*} )$，可得

$b_{n + 2} \cdot c_{n + 1} = b_{n} \cdot c_{n},$

两边同时乘以$b_{n + 1}$，可得

$b_{n + 1} b_{n + 2} c_{n + 1} = b_{n} b_{n + 1} \mathrm{c}_{\mathrm{n}},$

$\because b_{1} b_{2} c_{1} = b_{2} = 1 + d,$

$\therefore$数列$\left\{b_{n} b_{n + 1} c_{\mathrm{n}} \right\}$是一个常数列，且此常数为$1 {+} d$，

$\begin{array}{r l} & {b_{n} b_{n + 1} c_{n} = 1 + d,} \\ & {\therefore c_{n} = \frac{1 + d}{b_{n} b_{n + 1}} = \frac{1 + d}{d} \cdot \frac{d}{b_{n} b_{n + 1}} = (1 + \frac{1}{d}) \cdot \frac{b_{n + 1} - b_{n}}{b_{n} b_{n + 1}} = (1 + \frac{1}{d}) (\frac{1}{b_{n}} - \frac{1}{b_{n + 1}}),} \\ & {\therefore c_{1} + c_{2} + \dots + c_{n}} \\ & {= (1 + \frac{1}{d}) (\frac{1}{b_{1}} - \frac{1}{b_{2}}) + (1 + \frac{1}{d}) (\frac{1}{b_{2}} - \frac{1}{b_{3}}) + \dots + (1 + \frac{1}{d}) (\frac{1}{b_{n}} - \frac{1}{b_{n + 1}})} \\ & {= (1 + \frac{1}{d}) (\frac{1}{b_{1}} - \frac{1}{b_{2}} + \frac{1}{b_{2}} - \frac{1}{b_{3}} + \dots + \frac{1}{b_{n}} - \frac{1}{b_{n + 1}})} \\ & {= (1 + \frac{1}{d}) (\frac{1}{b_{1}} - \frac{1}{b_{n + 1}})} \\ & {= (1 + \frac{1}{d}) (1 - \frac{1}{b_{n + 1}})} \\ & {< 1 + \frac{1}{d},} \\ & {\therefore c_{1} + c_{2} + \dots + c_{n} < 1 + \frac{1}{d}, \text{故得证}.} \end{array}$