《2026年南方凤凰台5A新考案数学二轮提高版》

1. 求证：$\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+·s+\frac{1}{4n^{2}}＜\frac{1}{2}-\frac{1}{4n}(n\geq2,n\in\mathbf{N}^{*})$。

答案： 1.【解答】因为 $n \geq 2, n \in \mathbf{N}^*$ 时，$\frac{1}{n^2} ＜ \frac{1}{n(n - 1)} = \frac{1}{n - 1} - \frac{1}{n}$，所以
$\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + ·s + \frac{1}{4n^2} ＜ \frac{1}{4} \left( 1^2 + \frac{1}{2^2} + \frac{1}{3^2} + ·s + \frac{1}{n^2} \right) ＜ \frac{1}{4} \left( 1^2 + 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ·s + \frac{1}{n - 1} - \frac{1}{n} \right) = \frac{1}{4} \left( 2 - \frac{1}{n} \right) = \frac{1}{2} - \frac{1}{4n}$，
所以 $\frac{1}{4} + \frac{1}{16} + \frac{1}{36} + ·s + \frac{1}{4n^2} ＜ \frac{1}{2} - \frac{1}{4n} \quad (n \geq 2, n \in \mathbf{N}^*)$。

2. 设数列$\{ a_{n}\}$的前$n$项积$T_{n}=\frac{1}{2^{n + 1}-1}$，前$n$项和为$S_{n}$，求证：$S_{n}＞\frac{n}{2}-\frac{1}{2}+(\frac{1}{2})^{n + 1}$。

答案： 2.【解答】由 $T_n = \frac{1}{2^{n + 1} - 1}$，可得当 $n \geq 2$ 时，$a_n = \frac{T_n}{T_{n - 1}} = \frac{2^n - 1}{2^{n + 1} - 1}$，又当 $n = 1$ 时，$a_1 = T_1 = \frac{1}{3}$，也满足此式，所以 $a_n = \frac{2^n - 1}{2^{n + 1} - 1} ＞ \frac{2^n - 1}{2^{n + 1}} = \frac{1}{2} - \frac{1}{2^{n + 1}}$，所以 $S_n ＞ \left( \frac{1}{2} - \frac{1}{2^2} \right) + \left( \frac{1}{2} - \frac{1}{2^3} \right) + ·s + \left( \frac{1}{2} - \frac{1}{2^{n + 1}} \right) = \frac{n}{2} - \frac{\frac{1}{4} \left[ 1 - \left( \frac{1}{2} \right)^n \right]}{1 - \frac{1}{2}} = \frac{n}{2} - \frac{1}{2} + \left( \frac{1}{2} \right)^{n + 1}$。

例 1 - 1 已知数列$\{ a_{n}\}$的前$n$项和为$S_{n}$，$a_{1}=2$，$(n - 2)S_{n + 1}+2a_{n + 1}=nS_{n}$，$n\in\mathbf{N}^{*}$。
（1）求数列$\{ a_{n}\}$的通项公式；
（2）求证：$\frac{1}{a_{1}^{2}}+\frac{1}{a_{2}^{2}}+·s+\frac{1}{a_{n}^{2}}＜\frac{7}{16}$。

答案： 例1-1 【解答】
(1) 由 $(n - 2)S_{n + 1} + 2a_{n + 1} = nS_n$，得 $(n - 2) · S_{n + 1} + 2(S_{n + 1} - S_n) = nS_n$，整理得 $nS_{n + 1} = (n + 2)S_n$，故 $\frac{S_{n + 1}}{(n + 1)(n + 2)} = \frac{S_n}{n(n + 1)}$，所以 $\left\{ \frac{S_n}{n(n + 1)} \right\}$ 是常数列，且 $\frac{S_n}{n(n + 1)} = \frac{S_1}{1 × 2} = 1$，故 $S_n = n(n + 1)$。当 $n \geq 2$ 时，$a_n = S_n - S_{n - 1} = n(n + 1) - n(n - 1) = 2n$，当 $n = 1$ 时也满足此式，故 $a_n = 2n$。
(2) 方法一：$\frac{1}{a_n^2} = \frac{1}{4n^2} ＜ \frac{1}{4n^2 - 1} = \frac{1}{2} \left( \frac{1}{2n - 1} - \frac{1}{2n + 1} \right)$，故当 $n \geq 2$
时，$\frac{1}{a_1^2} + \frac{1}{a_2^2} + ·s + \frac{1}{a_n^2} ＜ \frac{1}{4} + \frac{1}{2} \left( \frac{1}{3} + \frac{1}{5} - \frac{1}{2n - 1} - \frac{1}{2n + 1} \right) = \frac{1}{4} + \frac{1}{2} \left( \frac{1}{3} - \frac{1}{2n - 1} \right) - \frac{1}{2} \left( \frac{1}{5} + \frac{1}{2n + 1} \right) = \frac{5}{12} - \frac{1}{2} \left( \frac{1}{2n - 1} + \frac{1}{2n + 1} \right) ＜ \frac{5}{12} ＜ \frac{7}{16}$。又 $\frac{1}{a_1^2} = \frac{1}{4} ＜ \frac{7}{16}$，故 $\frac{1}{a_1^2} + \frac{1}{a_2^2} + ·s + \frac{1}{a_n^2} ＜ \frac{7}{16}$。
方法二：当 $n \geq 3$ 时，$\frac{1}{a_n^2} = \frac{1}{4n^2} ＜ \frac{1}{4} × \frac{1}{n(n - 1)} = \frac{1}{4} \left( \frac{1}{n - 1} - \frac{1}{n} \right)$，所以 $\frac{1}{a_1^2} + \frac{1}{a_2^2} + ·s + \frac{1}{a_n^2} ＜ \frac{5}{16} + \frac{1}{4} \left( \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + ·s + \frac{1}{n - 1} - \frac{1}{n} \right) = \frac{5}{16} + \frac{1}{4} \left( \frac{1}{2} - \frac{1}{n} \right) = \frac{7}{16} - \frac{1}{4n} ＜ \frac{7}{16}$。当 $n = 1, 2$ 时也成立，所以 $\frac{1}{a_1^2} + \frac{1}{a_2^2} + ·s + \frac{1}{a_n^2} ＜ \frac{7}{16}$。
方法三：由于 $\frac{1}{n^2} ＜ \frac{1}{n^2 - 1} (n \geq 2)$，当 $n \geq 2$ 时，$\frac{1}{a_n^2} = \frac{1}{4n^2} ＜ \frac{1}{4} × \frac{1}{n^2 - 1} = \frac{1}{4} × \frac{1}{2} \left( \frac{1}{n - 1} - \frac{1}{n + 1} \right)$，所以 $\frac{1}{a_1^2} + \frac{1}{a_2^2} + ·s + \frac{1}{a_n^2} ＜ \frac{1}{4} + \frac{1}{4} × \frac{1}{2} \left( 1 - \frac{1}{3} + \frac{1}{2} + \frac{1}{4} - \frac{1}{3} - \frac{1}{5} + ·s + \frac{1}{n - 1} - \frac{1}{n + 1} \right) = \frac{1}{4} + \frac{1}{8} \left( 1 + \frac{1}{2} - \frac{1}{n} - \frac{1}{n + 1} \right) = \frac{7}{16} - \frac{1}{8} \left( \frac{1}{n} + \frac{1}{n + 1} \right) ＜ \frac{7}{16}$ 当 $n = 1$ 时也成立，所以 $\frac{1}{a_1^2} + \frac{1}{a_2^2} + ·s + \frac{1}{a_n^2} ＜ \frac{7}{16}$。