《2025年小题狂做高中数学选择性必修第三册人教版》

10. 教材变式$(1+\sqrt{2})^{5}+(1-\sqrt{2})^{5}=$82.

答案： 10. $82$ $(1+\sqrt{2})^{5}=C_{5}^{0}1^{5}(\sqrt{2})^{0}+C_{5}^{1}1^{4}(\sqrt{2})^{1}+C_{5}^{2}1^{3}(\sqrt{2})^{2}+C_{5}^{3}1^{2}(\sqrt{2})^{3}+C_{5}^{4}1^{1}(\sqrt{2})^{4}+C_{5}^{5}1^{0}(\sqrt{2})^{5}$，$(1-\sqrt{2})^{5}=C_{5}^{0}1^{5}(-\sqrt{2})^{0}+C_{5}^{1}1^{4}(-\sqrt{2})^{1}+C_{5}^{2}1^{3}(-\sqrt{2})^{2}+C_{5}^{3}1^{2}(-\sqrt{2})^{3}+C_{5}^{4}1^{1}(-\sqrt{2})^{4}+C_{5}^{5}1^{0}(-\sqrt{2)^{5}}$，可得两式和为$82$.

11. 若$x^{4}+(x+1)^{7}=a_{0}+a_{1}(x+2)+a_{2}(x+2)^{2}+·s+a_{7}(x+2)^{7}$，则$a_{3}=$27.

答案： 11. $27$ 由题意得$x^{4}+(x + 1)^{7}=[(x + 2)-2]^{4}+[(x + 2)-1]^{7}=a_{0}+a_{1}(x + 2)+a_{2}(x + 2)^{2}+·s+a_{7}(x + 2)^{7}$，则$a_{3}=C_{4}^{1}(-2)^{1}+C_{7}^{4}(-1)^{4}=-8 + 35 = 27$.

12. 观察下列等式：
$C_{5}^{1}+C_{5}^{5}=2^{3}-2$，
$C_{9}^{1}+C_{9}^{5}+C_{9}^{9}=2^{7}+2^{3}$，
$C_{13}^{1}+C_{13}^{5}+C_{13}^{9}+C_{13}^{13}=2^{11}-2^{5}$，
$C_{17}^{1}+C_{17}^{5}+C_{17}^{9}+C_{17}^{13}+C_{17}^{17}=2^{15}+2^{7}$，
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由以上等式推出一个一般的结论：
$\forall n\in N^{*},C_{4n+1}^{1}+C_{4n+1}^{5}+C_{4n+1}^{9}+·s+C_{4n+1}^{4n+1}=$$2^{4n - 1}+(-1)^{n}2^{2n - 1}$

答案： 12. $2^{4n - 1}+(-1)^{n}2^{2n - 1}$ 由等式规律知，等式右边第二项有$(-1)^{n}$，第一项与第二项中$2$的指数分别为$4n - 1$，$2n - 1$，因此$n\in\mathbf{N}^{*}$，$C_{4n + 1}^{0}+C_{4n + 1}^{1}+C_{4n + 1}^{2}+·s+C_{4n + 1}^{4n + 1}=2^{4n - 1}+(-1)^{n}2^{2n - 1}$.

13. 已知$(1-2x)^{5}=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}$，求下列各式的值.
(1)$a_{1}+a_{3}+a_{5}$；
(2)$(a_{0}+a_{2}+a_{4})^{2}-(a_{1}+a_{3}+a_{5})^{2}$；
(3)$|a_{0}|+|a_{1}|+|a_{2}|+|a_{3}|+|a_{4}|+|a_{5}|$；
(4)$a_{1}+2a_{2}+3a_{3}+4a_{4}+5a_{5}$.

答案： 13. 解：令$x=-1$，得$a_{0}-a_{1}+a_{2}-a_{3}+a_{4}-a_{5}=3^{5}$ ①，令$x = 1$，得$a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+a_{5}=-1$ ②.
(1)由① - ②并整理，得$a_{1}+a_{3}+a_{5}=-\frac{3^{5}+1}{2}$
(2)$(a_{0}+a_{2}+a_{4})^{2}-(a_{1}+a_{3}+a_{5})^{2}=(a_{0}+a_{1}+·s+a_{5})(a_{0}-a_{1}+·s-a_{5})=-3^{5}$.
(3)因为$(1 - 2x)^{5}$的展开式的各项系数的绝对值之和与$(1 + 2x)^{5}$的展开式的各项系数之和相等，所以在$(1 + 2x)^{5}$中，令$x = 1$，得$\vert a_{0}\vert+\vert a_{1}\vert+\vert a_{2}\vert+\vert a_{3}\vert+\vert a_{4}\vert+\vert a_{5}\vert=(1 + 2)^{5}=3^{5}$.
(4)由$(1 - 2x)^{5}=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}$两边求导，得$a_{1}+2a_{2}x+3a_{3}x^{2}+4a_{4}x^{3}+5a_{5}x^{4}=(-2)×5×(1 - 2x)^{4}$，令$x = 1$，得$a_{1}+2a_{2}+3a_{3}+4a_{4}+5a_{5}=-10×(1 - 2)^{4}=-10$.

14. 已知$f(x)=(1+2x)^{6}$.
(1) 若$(1-x)f(x)=a_{0}+a_{1}x+a_{2}x^{2}+·s+a_{7}x^{7}$，求$a_{7}$和$\sum_{i=1}^{7}a_{i}$的值；
(2) 求$f(x)$的展开式中系数最大的项.

答案： 14. 解：
(1)$f(x)=(1 + 2x)^{6}$的展开式的通项为$T_{r + 1}=C_{6}^{r}(2x)^{r}$，则$(1 - x)f(x)$的展开式中含$x^{7}$的项为$(-x)· C_{6}^{r}(2x)^{r}=-64x^{7}$，故$a_{7}=-64$.在$(1 - x)f(x)=a_{0}+a_{1}x+a_{2}x^{2}+·s+a_{7}x^{7}$中，令$x = 1$，得$0=a_{0}+a_{1}+a_{2}+·s+a_{7}$，令$x = 0$，得$1=a_{0}$，则$\sum_{i = 1}^{7}a_{i}=\sum_{i = 0}^{7}a_{i}-a_{0}=0 - 1=-1$.
(2)设展开式中第$(r + 1)$项的系数$C_{6}^{r}2^{r}=A_{r + 1}$，其中最大值为$A_{k}$，则$\begin{cases}A_{k}\geq A_{k - 1}\\A_{k}\geq A_{k + 1}\end{cases}$得$\begin{cases}C_{6}^{k - 1}2^{k - 1}\geq C_{6}^{k - 2}2^{k - 2}\\C_{6}^{k - 1}2^{k - 1}\geq C_{6}^{k}2^{k}\end{cases}$，即$\frac{6!}{(k - 1)!(7 - k)!}×2^{k - 1}\geq\frac{6!}{(k - 2)!(8 - k)!}×2^{k - 2}$，$\frac{6!}{(k - 1)!(7 - k)!}×2^{k - 1}\geq\frac{6!}{k!(6 - k)!}×2^{k}$，解得$\frac{14}{3}\leq k\leq\frac{17}{3}$，因为$k\in\mathbf{N}^{*}$，又$A_{1}=C_{6}^{0}2^{0}=1$，$A_{7}=C_{6}^{6}2^{6}=64$，所以当$r + 1 = k = 5$时，$A_{4}$取最大值$240$.因此$f(x)$的展开式中系数最大的项为$240x^{4}$.