《2026年南方凤凰台5A新考案数学二轮提高版》

例 1 - 2 （2025·宜昌模拟）已知$S_{n}$为数列$\{ a_{n}\}$的前$n$项和，且满足$a_{1}=1$，$3S_{n}=(n + 2)a_{n}$。
（1）求数列$\{ a_{n}\}$的通项公式；
（2）求证：$\frac{1}{a_{2}}+\frac{1}{a_{4}}+\frac{1}{a_{8}}+·s+\frac{1}{a_{2^{n}}}＜\frac{1}{2}$。（1）【解答】由 $3S_n = (n + 2)a_n$，得当 $n \geq 2$ 时，$3S_{n - 1} = (n + 1)a_{n - 1}$，两式相减得 $3a_n = (n + 2)a_n - (n + 1)a_{n - 1}$，整理得 $\frac{a_n}{a_{n - 1}} = \frac{n + 1}{n - 1}$，所以当 $n \geq 2$ 时，$a_n = a_1 × \frac{a_2}{a_1} × \frac{a_3}{a_2} × ·s × \frac{a_n}{a_{n - 1}} = 1 × \frac{3}{1} × \frac{4}{2} × ·s × \frac{n}{n - 2} × \frac{n + 1}{n - 1} = \frac{n(n + 1)}{2}$，又 $a_1 = 1$ 也满足此式，所以 $a_n = \frac{n(n + 1)}{2} (n \in \mathbf{N}^*)$。（2）【解答】因为 $\frac{1}{a_{2^n}} = \frac{2}{2^n(2^n + 1)} = \frac{1}{2^{n - 1}(2^n + 1)} = \frac{1}{2^n × 2^{n - 1}} - \frac{1}{2^{2n - 1}(2^n + 1)}$，所以 $\frac{1}{a_2} + \frac{1}{a_4} + \frac{1}{a_8} + ·s + \frac{1}{a_{2^n}} \leq \frac{1}{3} + \frac{1}{2^3} + \frac{1}{2^5} + ·s + \frac{1}{2^{2n - 1}} = \frac{\frac{1}{3} \left( 1 - \frac{1}{4^n} \right)}{1 - \frac{1}{4}} = \frac{1}{3} + \frac{1}{6} \left( 1 - \frac{1}{4^{n - 1}} \right) ＜ \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$，从而得 $\frac{1}{a_2} + \frac{1}{a_4} + \frac{1}{a_8} + ·s + \frac{1}{a_{2^n}} ＜ \frac{1}{2}$。

答案： （1）【解答】由 $3S_n = (n + 2)a_n$，得当 $n \geq 2$ 时，$3S_{n - 1} = (n + 1)a_{n - 1}$，两式相减得 $3a_n = (n + 2)a_n - (n + 1)a_{n - 1}$，整理得 $\frac{a_n}{a_{n - 1}} = \frac{n + 1}{n - 1}$，所以当 $n \geq 2$ 时，$a_n = a_1 × \frac{a_2}{a_1} × \frac{a_3}{a_2} × ·s × \frac{a_n}{a_{n - 1}} = 1 × \frac{3}{1} × \frac{4}{2} × ·s × \frac{n}{n - 2} × \frac{n + 1}{n - 1} = \frac{n(n + 1)}{2}$，又 $a_1 = 1$ 也满足此式，所以 $a_n = \frac{n(n + 1)}{2} (n \in \mathbf{N}^*)$。
（2）【解答】因为 $\frac{1}{a_{2^n}} = \frac{2}{2^n(2^n + 1)} = \frac{1}{2^{n - 1}(2^n + 1)} = \frac{1}{2^n × 2^{n - 1}} - \frac{1}{2^{2n - 1}(2^n + 1)}$，所以 $\frac{1}{a_2} + \frac{1}{a_4} + \frac{1}{a_8} + ·s + \frac{1}{a_{2^n}} \leq \frac{1}{3} + \frac{1}{2^3} + \frac{1}{2^5} + ·s + \frac{1}{2^{2n - 1}} = \frac{\frac{1}{3} \left( 1 - \frac{1}{4^n} \right)}{1 - \frac{1}{4}} = \frac{1}{3} + \frac{1}{6} \left( 1 - \frac{1}{4^{n - 1}} \right) ＜ \frac{1}{3} + \frac{1}{6} = \frac{1}{2}$，从而得 $\frac{1}{a_2} + \frac{1}{a_4} + \frac{1}{a_8} + ·s + \frac{1}{a_{2^n}} ＜ \frac{1}{2}$。

变式 1 （2025·台州质检）已知数列$\{ a_{n}\}$和$\{ b_{n}\}$满足$a_{1}(b_{1}+1)+a_{2}(b_{2}+1)+·s+a_{n}(b_{n}+1)=(2n - 3)·2^{n + 1}+6$，$n\in\mathbf{N}^{*}$，且$a_{1}=b_{1}=1$，$b_{n + 1}=2b_{n}+1$。
（1）求数列$\{ a_{n}\}$和$\{ b_{n}\}$的通项公式；
（2）求$[\sum_{n = 1}^{50}\frac{1}{\sqrt{a_{n}}}] $的值（其中$[x]$表示不大于$x$的最大整数，如$[3.2]=3$）。（1）【解答】由 $b_{n + 1} = 2b_n + 1$ 可得 $b_{n + 1} + 1 = 2(b_n + 1)$，且 $b_1 + 1 = 2$，所以数列 $\{ b_n + 1 \}$ 是首项和公比都为 2 的等比数列，所以 $b_n + 1 = 2 · 2^{n - 1} = 2^n$，故 $b_n = 2^n - 1$。由 $a_1(b_1 + 1) + a_2(b_2 + 1) + ·s + a_n(b_n + 1) = (2n - 3) · 2^{n + 1} + 6$，①，当 $n \geq 2$ 时，有 $2a_1 + 2^2a_2 + 2^3a_3 + ·s + 2^n a_n = (2n - 5) · 2^n + 6$，②，① - ②得 $2^n a_n = (2n - 1) · 2^n$，解得 $a_n = 2n - 1$，$a_1 = 1$ 也满足 $a_n = 2n - 1$，故对任意的 $n \in \mathbf{N}^*$，$a_n = 2n - 1$。（2）【解答】因为 $\frac{1}{a_n} = \frac{1}{\sqrt{2n - 1}} ＞ \frac{2}{\sqrt{2n - 1} + \sqrt{2n + 1}} = \frac{2(\sqrt{2n + 1} - \sqrt{2n - 1})}{(\sqrt{2n + 1} - \sqrt{2n - 1})(\sqrt{2n + 1} + \sqrt{2n - 1})} = - \sqrt{2n - 1} + \sqrt{2n + 1}$，所以 $\sum_{n = 1}^{50} \frac{1}{\sqrt{a_n}} ＞ (-1 + \sqrt{3}) + (- \sqrt{3} + \sqrt{5}) + (- \sqrt{5} + \sqrt{7}) + ·s + (- \sqrt{97} + \sqrt{99}) + (- \sqrt{99} + \sqrt{101}) = \sqrt{101} - 1$。另一方面，当 $n \geq 2$ 时，$\frac{1}{a_n} = \frac{1}{\sqrt{2n - 1}} = \frac{2}{2\sqrt{2n - 1}} ＜ \frac{2}{\sqrt{2n - 1} + \sqrt{2n - 3}} = \frac{2(\sqrt{2n - 1} - \sqrt{2n - 3})}{(\sqrt{2n - 1} + \sqrt{2n - 3})(\sqrt{2n - 1} - \sqrt{2n - 3})} = - \sqrt{2n - 3} + \sqrt{2n - 1}$，所以 $\sum_{n = 1}^{50} \frac{1}{\sqrt{a_n}} ＜ 1 + (-1 + \sqrt{3}) + (- \sqrt{3} + \sqrt{5}) + ·s + (- \sqrt{97} + \sqrt{99}) = \sqrt{99}$，所以 $\sqrt{101} - 1 ＜ \sum_{n = 1}^{50} \frac{1}{\sqrt{a_n}} ＜ \sqrt{99}$，又 $\sqrt{101} - 1 ＞ \sqrt{100} - 1 = 9$，$\sqrt{99} ＜ \sqrt{100} ＜ 10$，所以 $\left[ \sum_{n = 1}^{50} \frac{1}{\sqrt{a_n}} \right] = 9$。

答案： （1）【解答】由 $b_{n + 1} = 2b_n + 1$ 可得 $b_{n + 1} + 1 = 2(b_n + 1)$，且 $b_1 + 1 = 2$，所以数列 $\{ b_n + 1 \}$ 是首项和公比都为 2 的等比数列，所以 $b_n + 1 = 2 · 2^{n - 1} = 2^n$，故 $b_n = 2^n - 1$。由 $a_1(b_1 + 1) + a_2(b_2 + 1) + ·s + a_n(b_n + 1) = (2n - 3) · 2^{n + 1} + 6$，①，当 $n \geq 2$ 时，有 $2a_1 + 2^2a_2 + 2^3a_3 + ·s + 2^n a_n = (2n - 5) · 2^n + 6$，②，① - ②得 $2^n a_n = (2n - 1) · 2^n$，解得 $a_n = 2n - 1$，$a_1 = 1$ 也满足 $a_n = 2n - 1$，故对任意的 $n \in \mathbf{N}^*$，$a_n = 2n - 1$。
（2）【解答】因为 $\frac{1}{a_n} = \frac{1}{\sqrt{2n - 1}} ＞ \frac{2}{\sqrt{2n - 1} + \sqrt{2n + 1}} = \frac{2(\sqrt{2n + 1} - \sqrt{2n - 1})}{(\sqrt{2n + 1} - \sqrt{2n - 1})(\sqrt{2n + 1} + \sqrt{2n - 1})} = - \sqrt{2n - 1} + \sqrt{2n + 1}$，所以 $\sum_{n = 1}^{50} \frac{1}{\sqrt{a_n}} ＞ (-1 + \sqrt{3}) + (- \sqrt{3} + \sqrt{5}) + (- \sqrt{5} + \sqrt{7}) + ·s + (- \sqrt{97} + \sqrt{99}) + (- \sqrt{99} + \sqrt{101}) = \sqrt{101} - 1$。另一方面，当 $n \geq 2$ 时，$\frac{1}{a_n} = \frac{1}{\sqrt{2n - 1}} = \frac{2}{2\sqrt{2n - 1}} ＜ \frac{2}{\sqrt{2n - 1} + \sqrt{2n - 3}} = \frac{2(\sqrt{2n - 1} - \sqrt{2n - 3})}{(\sqrt{2n - 1} + \sqrt{2n - 3})(\sqrt{2n - 1} - \sqrt{2n - 3})} = - \sqrt{2n - 3} + \sqrt{2n - 1}$，所以 $\sum_{n = 1}^{50} \frac{1}{\sqrt{a_n}} ＜ 1 + (-1 + \sqrt{3}) + (- \sqrt{3} + \sqrt{5}) + ·s + (- \sqrt{97} + \sqrt{99}) = \sqrt{99}$，所以 $\sqrt{101} - 1 ＜ \sum_{n = 1}^{50} \frac{1}{\sqrt{a_n}} ＜ \sqrt{99}$，又 $\sqrt{101} - 1 ＞ \sqrt{100} - 1 = 9$，$\sqrt{99} ＜ \sqrt{100} ＜ 10$，所以 $\left[ \sum_{n = 1}^{50} \frac{1}{\sqrt{a_n}} \right] = 9$。