《2025年学霸高考黑题数学人教版》

变式训练 1. (2025·江苏八市三模)已知数列$\{ a_{n}\}$是等差数列，记其前$n$项和为$S_{n}$，且$S_{3}=a_{5}$，$a_{2n}=2a_{n}+\dfrac {1}{4}$。
(1)求数列$\{ a_{n}\}$的通项公式；
(2)将数列$\{ a_{n}\}$与$\{ \sqrt {S_{n}}\}$的所有项从小到大排列得到数列$\{ b_{n}\}$。
①求$\{ b_{n}\}$的前$20$项和；
②证明：$\dfrac {1}{b_{1}^{2}}+\dfrac {1}{b_{2}^{2}}+·s +\dfrac {1}{b_{n}^{2}} ＜ 32$。

答案： 变式训练1.
(1)解：设等差数列$\{a_n\}$的公差为$d$，由$S_3 = a_5$，得$3a_1 + 3d = a_1 + 4d$，即$2a_1 = d$，由$a_{2n}=2a_{n}+\dfrac {1}{4}$，取$n = 1$，得$a_2 = 2a_1 + \frac{1}{4} = a_1 + d$，即$a_1 = d - \frac{1}{4}$，解得$a_1 = \frac{1}{4}$，$d = \frac{1}{2}$，所以$a_n = \frac{1}{2}n - \frac{1}{4}$。
(2)①解：由
(1)知，$S_n = \frac{1}{4}n + \frac{n(n - 1)}{2} × \frac{1}{2} = \frac{1}{4}n^2$，所以$\sqrt{S_n} = \frac{1}{2}n = \frac{2n}{4}$。因为$a_n = \frac{1}{2}n - \frac{1}{4} = \frac{2n - 1}{4}$，所以$b_n = \frac{1}{4}n$，所以$\{b_n\}$的前20项和为$20 × \frac{1}{4} + \frac{20 × 19}{2} × \frac{1}{4} = \frac{105}{2}$。
②证明：因为$b_n^2 = \frac{1}{16}n^2$，所以$\frac{1}{b_n^2} = \frac{16}{n^2} ＜ \frac{16}{n(n - 1)} = 16(\frac{1}{n - 1} - \frac{1}{n})(n \geq 2)$，当$n = 1$时，$\frac{1}{b_1^2} = 16 ＜ 32$；当$n \geq 2$时，$\frac{1}{b_1^2} + \frac{1}{b_2^2} + ·s + \frac{1}{b_n^2} ＜ 16 + 16[(\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + ·s + (\frac{1}{n - 1} - \frac{1}{n})] = 32 - \frac{16}{n} ＜ 32$。综上可

变式训练 2. (2025·江苏泰州模拟)设数列$\{ a_{n}\}$的前$n$项和为$S_{n}$，$2S_{n}=n^{2}+5n$。
(1)求$\{ a_{n}\}$的通项公式；
(2)设$b_{n}=n· 2^{n}$，求数列$\left\{ \dfrac {S_{n}+4}{a_{n}b_{n}}\right\}$的前$n$项和$T_{n}$；
(3)设$c_{n}=\dfrac {1}{\sqrt {a_{n}-2}}$，求证：$c_{1}+c_{2}+c_{3}+·s +c_{n} ＞ 2\sqrt {n}-\dfrac {3}{2}$。

答案： 变式训练2.
(1)解：$\because 2S_n = n^2 + 5n$，即$S_n = \frac{n^2 + 5n}{2}$，当$n = 1$时，$S_1 = \frac{1^2 + 5}{2} = 3$，$\therefore a_1 = S_1 = 3$；当$n \geq 2$时，$S_{n - 1} = \frac{(n - 1)^2 + 5(n - 1)}{2}$，$\therefore a_n = S_n - S_{n - 1} = \frac{n^2 + 5n}{2} - \frac{(n - 1)^2 + 5(n - 1)}{2} = n + 2$，而$a_1 = 3$也满足上式，$\therefore a_n = n + 2$。
(2)解：$\because S_n = \frac{n^2 + 5n}{2}$，$a_n = n + 2$，$b_n = n · 2^n$，
$\therefore \frac{S_n + 4}{a_n b_n} = \frac{\frac{n^2 + 5n}{2} + 4}{(n + 2) × n × 2^n} = \frac{n^2 + 5n + 8}{(n + 2) × n × 2^{n + 1}} = \frac{(n + 2)n + 3n + 8}{(n + 2) × n × 2^{n + 1}}$，
$\frac{1}{(n + 2) × n × 2^{n + 1}}[\frac{3n + 8}{(n + 2) × n} - \frac{1}{n} - \frac{1}{n + 1} - \frac{1}{n + 2}] = \frac{1}{4}[\frac{1}{n × 2^{n - 1}} - \frac{1}{(n + 1) × 2^n}]$，
$\therefore T_n = \frac{1}{4}[1 - (\frac{1}{2}) + \frac{1 - \frac{1}{2}}{1 × 2^0} - \frac{1}{3 × 2^2} + \frac{1}{2 × 2^1} - \frac{1}{4 × 2^3} + \frac{1}{3 × 2^2} - \frac{1}{5 × 2^4} + ·s +$
$\frac{1}{n × 2^{n - 1}} - \frac{1}{(n + 1) × 2^n}] = \frac{1}{4}[1 + \frac{\frac{1}{2}}{2} - \frac{1}{2} × \frac{\frac{1}{4}}{1 - \frac{1}{2}} + 1 + \frac{1}{4} - \frac{1}{(n + 1) · 2^n}]$，
$= \frac{1}{(n + 2) · 2^{n + 1}} = \frac{7}{4 × 2^3} - \frac{1}{(n + 1) · 2^n} · \frac{(n + 1) · 2^n}{(n + 2) · 2^{n + 1}}$，
(3)证明：由
(1)可得$c_n = \frac{1}{\sqrt{a_{n} - 2}} = \frac{1}{\sqrt{n}}$，因为$\frac{1}{\sqrt{k}} = \frac{2}{2\sqrt{k}} ＞ \frac{2}{\sqrt{k} + \sqrt{k + 1}} = 2(\sqrt{k + 1} - \sqrt{k})(k \in N^*)$，所以$c_1 + c_2 + ·s + c_n = 1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ·s + \frac{1}{\sqrt{n}} ＞ 2(\sqrt{2} - 1) + 2(\sqrt{3} - \sqrt{2}) + ·s + 2(\sqrt{n + 1} - \sqrt{n}) = 2\sqrt{n + 1} - 2$，当$n \geq 1$时，$2\sqrt{n + 1} - 2 - (2\sqrt{n} - \frac{3}{2}) = 2(\sqrt{n + 1} - \sqrt{n}) - \frac{1}{2} = \frac{2}{\sqrt{n + 1} + \sqrt{n}} - \frac{1}{2}$，当$n = 1$时，$2(\sqrt{2} - 1) - \frac{1}{2} = 2\sqrt{2} - \frac{5}{2} \approx 2.828 - 2.5 = 0.328 ＞ 0$；当$n = 2$时，$2(\sqrt{3} - \sqrt{2}) - \frac{1}{2} \approx 2(1.732 - 1.414) - 0.5 = 0.636 - 0.5 = 0.136 ＞ 0$；随着$n$增大，$\frac{2}{\sqrt{n + 1} + \sqrt{n}}$趋近于0，当$n$足够大时，$2\sqrt{n + 1} - 2 ＜ 2\sqrt{n} - \frac{3}{2}$，但原证明过程可能存在笔误，正确证法应为：$\frac{1}{\sqrt{k}} = \frac{2}{2\sqrt{k}} ＞ \frac{2}{\sqrt{k} + \sqrt{k - 1}} = 2(\sqrt{k} - \sqrt{k - 1})(k \geq 2)$，则$c_1 + c_2 + ·s + c_n = 1 + \sum_{k=2}^n \frac{1}{\sqrt{k}} ＞ 1 + 2\sum_{k=2}^n (\sqrt{k} - \sqrt{k - 1}) = 1 + 2(\sqrt{n} - 1) = 2\sqrt{n} - 1 ＞ 2\sqrt{n} - \frac{3}{2}$。