《2024年高职高考全真模拟试卷辽海出版社高中数学全一册人教版》

22. 在$\triangle ABC$中，$A = 120^{\circ}$，$AB = AC$，$BC = 4\sqrt{3}$.
(1)求$\triangle ABC$的面积；
(2)若$M$为$AC$的中点，求$BM$的长.

答案： 解：
(1)由$AB = AC$知，$\triangle ABC$是等腰三角形
故$B = C = 30^{\circ}$
由正弦定理知，$\frac{BC}{\sin A} = \frac{AC}{\sin B}$
即$\frac{4\sqrt{3}}{\sin120^{\circ}} = \frac{AC}{\sin30^{\circ}}$，得$AC = \frac{4\sqrt{3} \times \sin30^{\circ}}{\sin120^{\circ}} = 4$
由三角形面积公式得
$S_{\triangle ABC} = \frac{1}{2}AC \times BC \times \sin C = \frac{1}{2} \times 4 \times 4\sqrt{3} \times \sin30^{\circ} = 4\sqrt{3}$
(2)在$\triangle ABC$中，$M$为$AC$的中点，由余弦定理得
$BM^{2} = AB^{2} + AM^{2} - 2AB \cdot AM \cdot \cos A$
$= 4^{2} + 2^{2} - 2 \times 4 \times 2 \times \cos120^{\circ} = 28$
故$BM = 2\sqrt{7}$

23. 已知等差数列$\{a_{n}\}$满足：$a_{5} = 11$，$a_{2} + a_{9} = 24$，设数列$\{a_{n}\}$前$n$项和为$S_{n}$.
(1)求$a_{n}$及$S_{n}$；
(2)令$b_{n} = \frac{1}{a_{n}^{2} - 1}(n\in\mathbf{N}^{*})$，求数列$\{b_{n}\}$的前$n$项和$T_{n}$.

答案： 解：
(1)设等差数列$\{a_{n}\}$的首项为$a_{1}$，公差为$d$，
因为$a_{5} = 11$，$a_{2} + a_{9} = 24$，
所以$\begin{cases}a_{1} + 4d = 11\\2a_{1} + 9d = 24\end{cases}$，
解得$\begin{cases}a_{1} = 3\\d = 2\end{cases}$，
所以$a_{n} = 3 + 2(n - 1) = 2n + 1(n \in \mathbf{N}_{+})$
$S_{n} = 3n + \frac{n(n - 1)}{2} \times 2 = n^{2} + 2n(n \in \mathbf{N}_{+})$
(2)由
(1)知$a_{n} = 2n + 1$
所以$b_{n} = \frac{1}{a_{n}^{2} - 1} = \frac{1}{(2n + 1)^{2} - 1} = \frac{1}{4n^{2} + 4n} = \frac{1}{4} \cdot \frac{1}{n(n + 1)} = \frac{1}{4} \cdot (\frac{1}{n} - \frac{1}{n + 1})(n \in \mathbf{N}_{+})$
所以$T_{n} = \frac{1}{4} \cdot (1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} + \cdots + \frac{1}{n} - \frac{1}{n + 1}) = \frac{1}{4} \cdot (1 - \frac{1}{n + 1}) = \frac{n}{4(n + 1)}$
即数列$\{b_{n}\}$的前项和为$T_{n} = \frac{n}{4(n + 1)}(n \in \mathbf{N}_{+})$