《2026年南方凤凰台5A新考案高中数学二轮基础版》

(2)(2025·鹰潭一模)已知$(1 + 2x)^n = a_0 + a_1x + a_2x^2 + a_3x^3 + ·s + a_nx^n$，随机变量$\xi \sim N(1,\frac{1}{4})$，若$\frac{a_1}{a_2} = E(\xi)D(\xi)$，则$a_1 + a_2 + a_3 + ·s + a_n$的值为(B)

A.81
B.242
C.243
D.80

答案：
(2) B 【解析】因为随机变量$\xi \sim N(1,\frac{1}{4})$，所以$E(\xi) = 1$，$D(\xi) = \frac{1}{4}$. 因为$(1 + 2x)^{n} = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + ·s + a_{n}x^{n}$，所以$a_{1} = C_{n}^{1} · 2 = 2n$，$a_{2} = C_{n}^{2} · 2^{2} = 2n(n - 1)$，所以$\frac{a_{1}^{2}}{a_{2}} = \frac{2n}{2n(n - 1)} · \frac{1}{n - 1} = \frac{1}{4}$，解得$n = 5$. 令$f(x) = (1 + 2x)^{5} = a_{0} + a_{1}x + a_{2}x^{2} + a_{3}x^{3} + a_{4}x^{4} + a_{5}x^{5}$，则$f(0) = a_{0} = 1$，$f(1) = (1 + 2)^{5} = a_{0} + a_{1} + a_{2} + a_{3} + a_{4} + a_{5}$，故$a_{1} + a_{2} + a_{3} + a_{4} + a_{5} = 3^{5} - 1 = 242$.

变式 2 (1)(2025·枣庄模拟)已知$(3x + 1)^{2025} = a_0 + a_1x + a_2x^2 + ·s + a_{2025}x^{2025}$，则$\frac{4}{3}a_1 + (\frac{4}{3})^2a_2 + ·s + (\frac{4}{3})^{2025}a_{2025}$被 4 除的余数为(D)

A.3
B.2
C.1
D.0

答案：
(1) D 【解析】令$x = 0$，得$a_{0} = 1^{2025} = 1$，令$x = \frac{4}{3}$，可得$5^{2025} = a_{0} + \frac{4}{3}a_{1} + (\frac{4}{3})^{2}a_{2} + ·s + (\frac{4}{3})^{2025}a_{2025}$，所以$\frac{4}{3}a_{1} + (\frac{4}{3})^{2}a_{2} + ·s + (\frac{4}{3})^{2025}a_{2025} = 5^{2025} - 1$. 因为$5^{2025} = (4 + 1)^{2025} = C_{2025}^{0}4^{2025} · 1^{0} + C_{2025}^{1}4^{2024} · 1^{1} + ·s + C_{2025}^{2024}4^{1} · 1^{2024} + C_{2025}^{2025}4^{0} = C_{2025}^{1}4^{2024} + C_{2025}^{2}4^{2023} + ·s + C_{2025}^{2024}4^{1} + 1$，所以$5^{2025}$被4除的余数为1，即$5^{2025} - 1$被4除的余数为0.

(2)(多选)已知$(x^2 + x + 1)^9 = a_0 + a_1x + a_2x^2 + ·s + a_{18}x^{18}$，下列说法正确的有(AD)

A.$a_0 = 1$
B.$a_2 = 42$
C.$a_2 + a_4 + ·s + a_{18} = \frac{3^9 + 1}{2}$
D.$a_1 + 2a_2 + 3a_3 + ·s + 18a_{18} = 3^{11}$

答案：
(2) AD 【解析】对于A，令$x = 0$，得$a_{0} = 1$，故A正确. 对于B，$a_{2} = C_{3}^{2}C_{3}^{8} + C_{3}^{3}C_{3}^{3} = 9 + 36 = 45$，故B错误. 对于C，令$x = 1$，则$a_{0} + a_{1} + a_{2} + ·s + a_{18} = 3^{9}$；令$x = - 1$，则$a_{0} - a_{1} + a_{2} - ·s + a_{18} = 1$，两式相加得$2(a_{0} + a_{2} + ·s + a_{18}) = 3^{9} + 1$，又$a_{0} = 1$，所以$a_{2} + a_{4} + ·s + a_{18} = \frac{3^{9} + 1}{2} - 1 = \frac{3^{9} - 1}{2}$，故C错误. 对于D，因为$[(x^{2} + x + 1)^{9}]' = 9(x^{2} + x + 1)^{8}(2x + 1),(a_{0} + a_{1}x + a_{2}x^{2} + ·s + a_{18}x^{18})' = a_{1} + 2a_{2}x + ·s + 18a_{18}x^{17}$，所以$9(x^{2} + x + 1)^{8}(2x + 1) = a_{1} + 2a_{2}x + ·s + 18a_{18}x^{17}$，令$x = 1$，则$a_{1} + 2a_{2} + 3a_{3} + ·s + 18a_{18} = 9 × 3^{8} × 3 = 3^{11}$，故D正确.

例 3 已知$(\sqrt{x} - \frac{2}{x})^n$的展开式中只有第 5 项的二项式系数最大，则该展开式中各项系数的最小值为(C)

A.-448
B.-1024
C.-1792
D.-5376

答案： C 【解析】因为展开式中只有第5项的二项式系数最大，所以$n = 8$，则展开式的通项$T_{r + 1} = C_{8}^{r}(\sqrt{x})^{8 - r} · ( - \frac{2}{x})^{r} = ( - 2)^{r}C_{8}^{r}x^{\frac{8 - 3r}{2}},r = 0,1,·s,8$，于是该展开式中各项系数$a_{r} = ( - 2)^{r}C_{8}^{r},r = 0,1,·s,8$. 当系数取最小值时，$r$为奇数且$\begin{cases} ( - 2)^{r}C_{8}^{r} \leq ( - 2)^{r + 2}C_{8}^{r + 2} \\ ( - 2)^{r}C_{8}^{r} \leq ( - 2)^{r - 2}C_{8}^{r - 2} \end{cases}$即$\begin{cases} r \geq \frac{29}{4} \\ r \leq \frac{33}{4} \end{cases}$解得$r = 5$，所以最小值为$a_{5} = ( - 2)^{5}C_{8}^{5} = - 1792$。

变式 3 (2024·全国甲卷)$(\frac{1}{3} + x)^{10}$的展开式中，各项系数的最大值是5。

答案： 5【解析】展开式的通项$T_{r + 1} = C_{10}^{r}(\frac{1}{3})^{10 - r}x^{r},0 \leq r \leq 10$且$r \in Z$. 设展开式中第$r + 1$项系数最大，则$\begin{cases} C_{10}^{r}(\frac{1}{3})^{10 - r} \geq C_{10}^{r + 1}(\frac{1}{3})^{9 - r} \\ C_{10}^{r}(\frac{1}{3})^{10 - r} \geq C_{10}^{r - 1}(\frac{1}{3})^{11 - r} \end{cases}$可得$\begin{cases} r \geq \frac{29}{4} \\ r \leq \frac{33}{4} \end{cases}$，又$r \in Z$，所以$r = 8$. 故展开式中系数最大的项是第9项，且该项系数为$C_{10}^{8}(\frac{1}{3})^{2} = 5$。