浙江 2015 · 数学 q20

题面

20．（15分）（2015•浙江）已知数列$\{a_{\mathrm{n}} \}$满足$\mathrm{a}_{1} = \frac{1}{2}$且$a_{\mathrm{n} + 1} = a_{\mathrm{n}} - a_{\mathrm{n}}^{2} ( n \in \mathrm{N}^{*} )$

（1）证明：$1 \leq \frac{a_{n}}{a_{n + 1}} \leq 2 ( n \in N^{\ast} )$；

（2）设数列$\{a_{\mathrm{n}}^{2} \}$的前n项和为$\mathrm{S}_{\mathrm{n}}$，证明$\frac{1}{2 ( n + 2 )} \leqslant \frac{\mathrm{S_{n}}}{n} \leqslant \frac{1}{2 ( n + 1 )} \ ( n \in \mathrm{N}^{*} )$

2015年浙江省高考数学试卷（理科）

解析

分析

（1）通过题意易得$0 < \mathrm{a}_{\mathrm{n}} \leq \frac{1}{2} ( \mathrm{n} \in \mathrm{N}^{\ast} )$，利用$a_{\mathrm{n}} - a_{\mathrm{n} + 1} = a_{\mathrm{n}}^{2}$可得$\frac{a_{\mathrm{n}}}{a_{\mathrm{n} + 1}} \geq 1$，利用$\frac{a_{n}}{a_{n + 1}} = \frac{a_{n}}{a_{n} - a_{n}^{2}} = \frac{1}{1 - a_{n}} \leq 2$，即得结论；

（2）通过$a_{n}^{2} = a_{n} - a_{n + 1}$累加得$\mathrm{S_{n}} = \frac{1}{2} - \mathrm{a_{n + 1}}$，利用数学归纳法可证明$\frac{1}{1 + \mathrm{n}} \geq a_{\mathrm{n}} \geq \frac{1}{2 \mathrm{n}} \ ( \mathrm{n} \geq 2 )$，从而$\frac{\frac{1}{2} - \frac{1}{2 ( n + 1 )}}{n} \geq \frac{\frac{1}{2} - a_{n + 1}}{n} \geq \frac{\frac{1}{2} - \frac{1}{n + 2}}{n}$，化简即得结论

解答

证明：（1）由题意可知：$0 < a_{n} \leq \frac{1}{2} ( n \in N^{*} )$，

又$\because a_{2} = a_{1} - a_{1}^{2} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}$，$\therefore \frac{a_{1}}{a_{2}} = \frac{\frac{1}{2}}{\frac{1}{4}} = 2$，

又$\because a_{n} - a_{n + 1} = a_{n}^{2}$，$\therefore a_{n} > a_{n + 1}$，$\therefore \frac{a_{n}}{a_{n + 1}} \geq 1$，

$\therefore \frac{a_{n}}{a_{n + 1}} = \frac{a_{n}}{a_{n} - a_{n}^{2}} = \frac{1}{1 - a_{n}} \leq 2$，

$\therefore 1 \leq \frac{a_{n}}{a_{n + 1}} \leq 2 ( n \in N^{*} )$；

（2）由已知，$a_{n}^{2} = a_{n} - a_{n + 1} , \quad a_{n - 1}^{2} = a_{n - 1} - a_{n} , \quad \ldots , \quad a_{1}^{2} = a_{1} - a_{2}$，

累加，得$\mathrm{S_{n}} = a_{n}^{2} + a_{n - 1}^{2} + \ldots + a_{1}^{2} = \mathrm{a_{1}} - \mathrm{a_{n + 1}} = \frac{1}{2} - \mathrm{a_{n + 1}}$

易知当n=1时，要证式子显然成立；

当$n \ge 2$时，$\frac{S_{n}}{n} = \frac{\frac{1}{2} - a_{n + 1}}{n}$

下面证明：$\frac{1}{1 + \mathrm{n}} \geq a_{\mathrm{n}} \geq \frac{1}{2 \mathrm{n}} \ ( \mathrm{n} \geq 2 )$

易知当n=2时成立，假设当n=k时也成立，则$a_{k + 1} = - ( a_{k} - \frac{1}{2} )^{2} + \frac{1}{4}$，

由二次函数单调性知：$a_{n + 1} \geq - ( \frac{1}{2 \mathrm{k}} - \frac{1}{2} )^{2} + \frac{1}{4} = \frac{2 \mathrm{k} - 1}{4 \mathrm{k}^{2}} \geq \frac{1}{2 ( \mathrm{k} + 1 )}$，

$\mathrm{a}_{\mathrm{n} + 1} \leq - \left(\frac{1}{\mathrm{k} + 1} - \frac{1}{2}\right)^{2} + \frac{1}{4} = \frac{\mathrm{k}}{(\mathrm{k} + 1)^{2}} \leq \frac{1}{\mathrm{k} + 2},$

$\therefore \frac{1}{2 ( k + 1 )} \leq \frac{1}{a_{k + 1}} \leq \frac{1}{k + 2}$，即当n=k+1时仍然成立，

故对n≥2，均有$\frac{1}{1 + \mathrm{n}} \geq \mathrm{a}_{\mathrm{n}} \geq \frac{1}{2 \mathrm{n}}$，

$\therefore \frac{1}{2 (n + 1)} = \frac{\frac{1}{2} - \frac{1}{2 (n + 1)}}{n} \geq \frac{\frac{1}{2} - a_{n + 1}}{n} \geq \frac{\frac{1}{2} - \frac{1}{n + 2}}{n} = \frac{1}{2 (n + 2)},$

$\text{即} \frac{1}{2 (n + 2)} \leqslant \frac{S_{n}}{n} \leqslant \frac{1}{2 (n + 1)} \quad (n \in \mathbb{N}^{*}).$