浙江 2016 · 数学 q20

题面

20．（15分）（2016•浙江）设数列满足$| a_{\mathrm{n}} - \frac{\widetilde{a}_{\mathrm{n} + 1}}{2} | \leq 1 , \mathrm{n}{\in} \mathrm{N}^{*}$

（Ⅰ）求证：$\left| \mathrm{a}_{\mathrm{n}} | 2 2^{\mathrm{n}^{- 1}} \left( \left| \mathrm{a}_{1} \right| - 2 \right) \left( \mathrm{n}{\in} \mathrm{N}^{*} \right) \right.$

（Ⅱ）若$\lvert \mathrm{a_{n}} \rvert \leq ( \frac{3}{2} )^{\mathrm{n}} , \mathrm{n} \in \mathrm{N}^{*}$，证明：$| a_{\mathrm{n}} | {\leq} 2 , \ n {\in} N^{*}$

2016年浙江省高考数学试卷（理科）

解析

分析

（I）使用三角不等式得出$| a_{\mathrm{n}} | - \frac{1}{2} | a_{\mathrm{n} + 1} | {\le} 1$，变形得$\frac{\vert a_{n} \vert}{2^{n}} - \frac{\vert a_{n + 1} \vert}{2^{n + 1}} \leq \frac{1}{2^{n}} ,$使用累加法可求得$\frac{\mid{\bar{a}}_{1} \mid}{2} - \frac{\mid{\bar{a}}_{n} \mid}{2^{n}}{<} 1$，即结论成立；

（II）利用（I）的结论得出$\frac{\vert a_{\mathfrak{n}} \vert}{2^{\mathfrak{n}}} - \frac{\vert a_{\mathfrak{n}} \vert}{2^{\mathfrak{n}}} < \frac{1}{2^{\mathfrak{n} - 1}} ,$进而得出$\lvert \mathrm{a_{n}} \rvert{<} 2 + ( \frac{3}{4} )^{\ \mathrm{m}}{\bullet} 2^{\mathrm{n}}$，利用m的任意性可证$\left| a_{\mathrm{n}} \right| {\le} 2$

解答

解：（I）$\because \vert{\mathrm{a_{n}} - \frac{\bar{{a}}_{\mathrm{n + 1}}}{2} \vert \leq 1} , \cdot \vert{\mathrm{a_{n}} \vert - \frac{1}{2} \vert{\mathrm{a_{n + 1}} \vert \leq 1}}$

$\therefore \frac{\mid a_{n} \mid}{2^{n}} - \frac{\mid a_{n + 1} \mid}{2^{n + 1}} \leq \frac{1}{2^{n}}, n \in \mathrm{N}^{*},$

$\therefore \frac{\left| a_{1} \right|}{2} - \frac{\left| a_{n} \right|}{2^{n}} = \left(\frac{\left| a_{1} \right|}{2} - \frac{\left| a_{2} \right|}{2^{2}}\right) + \left(\frac{\left| a_{2} \right|}{2^{2}} - \frac{\left| a_{3} \right|}{2^{3}}\right) + \dots + \left(\frac{\left| a_{n - 1} \right|}{2^{n - 1}} - \frac{\left| a_{n} \right|}{2^{n}}\right)$

$\leq \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots + \frac{1}{2^{n}} = \frac{\frac{1}{2} (1 - \frac{1}{2^{n}})}{1 - \frac{1}{2}} = 1 - \frac{1}{2^{n}} < 1.$

$\therefore | a_{n} | \geq 2^{n^{- 1}} (| a_{1} | - 2) (n \in \mathbb{N}^{*}).$

（II）任取$\mathrm{\ n}{\in} \mathrm{N}^{\ast}$，由（I）知，对于任意$\mathfrak{m} > \mathfrak{n}$

$\frac{\left| a_{n} \right|}{2^{n}} - \frac{\left| a_{m} \right|}{2^{m}} = \left(\frac{\left| a_{n} \right|}{2^{n}} - \frac{\left| a_{n + 1} \right|}{2^{n + 1}}\right) + \left(\frac{\left| a_{n + 1} \right|}{2^{n + 1}} - \frac{\left| a_{n + 2} \right|}{2^{n + 2}}\right) + \dots + \left(\frac{\left| a_{m - 1} \right|}{2^{m - 1}} - \frac{\left| a_{m} \right|}{2^{m}}\right)$

$\leq \frac{1}{2^{n}} + \frac{1}{2^{n + 1}} + \dots + \frac{1}{2^{m - 1}} = \frac{\frac{1}{2^{n}} (1 - \frac{1}{2^{m - n + 1}})}{1 - \frac{1}{2}} < \frac{1}{2^{n - 1}}.$

\therefore | a_{n} | < \left(\frac{1}{2^{n - 1}} + \frac{| a_{m} |}{2^{m}}\right) \cdot 2^{n} \leq \left[\frac{1}{2^{n - 1}} + \frac{1}{2^{m}} \cdot \left(\frac{3}{2}\right)^{m} \right] \cdot 2^{n} = 2 + \left(\frac{3}{4}\right)^{m} \cdot 2^{n}.\tag{①}

由m的任意性可知$\left| a_{\mathrm{n}} \right| \leq 2$

否则，存在$\mathrm{n 0}{\in} \mathrm{N}^{\ast}$，使得$| a_{n_{0}} | > 2$，

取正整数$\mathrm{m}_{0}{>} \log_{\mathrm{\frac{3}{4}}} \frac{\vert a_{n_{0}} \vert - 2}{2^{n_{0}}}$且$\mathrm{m}_{0} > \mathrm{n}_{0}$，则

{2}^{\mathrm{n_{0}}} \cdot ( \frac{3}{4} )^{\quad \mathrm{n_{0}}} \cdot \mathrm{n_{0}} \cdot ( \frac{3}{4} )^{\quad \mathrm{}} \frac \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm{} \mathrm①式矛盾

综上，对于任意$\mathrm{{n}}{\in} \mathrm{{N}}^{\ast}$，都有$\left. a_{\mathrm{n}} \right. \leq 2$