di-zhang-fdu/OlympiadExtended huggingface dataset:
Subfield: Probability and Statistics


Question: 周末李梦提出和父亲、母亲、弟弟进行羽毛球比赛，李梦与他们三人各进行一场比赛，共进行三场比赛，而且三场比赛相互独立.根据李梦最近分别与父亲、母亲、弟弟比赛的情况，得到如下统计表：

|    | 父亲 | 母亲 | 弟弟 |
|:---------:|:------:|:------:|:-------:|
|比赛的次数 |  50   |  60   |   40   |
|李梦获胜的次数 |  10   |  30   |   32   |

以上表中的频率作为概率，求解下列问题。
如果按照第一场与父亲比赛、第二场与母亲比赛、第三场与弟弟比赛的顺序进行比赛: 如果李梦胜一场得1分，负一场得0分，设李梦的得分为X，求X的分布列与期望.

Solution:
$X的所有可能取值为0，1，2，3，$
$则 P(X=0)=\left(1-\frac{1}{5}\right) \times \left(1-\frac{1}{2}\right) \times \left(1-\frac{4}{5}\right) = \frac{2}{25}，$
$P(X=1)=\left(1-\frac{1}{5}\right) \times \left(1-\frac{1}{2}\right) \times \frac{4}{5} + \left(1-\frac{1}{5}\right) \times \frac{1}{2} \times \left(1-\frac{4}{5}\right) + \frac{1}{5} \times \left(1-\frac{1}{2}\right) \times \left(1-\frac{4}{5}\right) = \frac{21}{50}，$
$P(X=2)=\left(1-\frac{1}{5}\right) \times \frac{1}{2} \times \frac{4}{5} + \frac{1}{5} \times \frac{1}{2} \times \left(1-\frac{4}{5}\right) + \frac{1}{5} \times \left(1-\frac{1}{2}\right) \times \frac{4}{5} = \frac{21}{50}，$
$P(X=3)= \frac{1}{5} \times \frac{1}{2} \times \frac{4}{5} = \frac{2}{25}.$
故分布列为

| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P | $\frac{2}{25}$ | $\frac{21}{50}$ | $\frac{21}{50}$ | $\frac{2}{25}$ |

$E(X)=0\times \frac{2}{25} +1\times \frac{21}{50} +2\times \frac{21}{50} +3\times \frac{2}{25} = \frac{3}{2}.$

Final answer: 分布列为

| X | 0 | 1 | 2 | 3 |
|---|---|---|---|---|
| P | $\frac{2}{25}$ | $\frac{21}{50}$ | $\frac{21}{50}$ | $\frac{2}{25}$ |

$期望为\frac{3}{2}$
Subfield: Conic Sections


Question: $已知抛物线C:y^2=2px(p>0)过点M(1，-2\sqrt{2})，直线l经过抛物线的焦点F与抛物线交于A，B两点.$

<img_2277>
$若直线OA，OB的斜率分别为k_1，k_2，且k_1+k_2=2，求直线l的方程.$

Solution:
$由题意知直线l的斜率存在且不为0，$

$设直线l的方程为y=k(x-2)，A(x_1,y_1)，B(x_2,y_2)，$

联立
$\left\{\begin{matrix}y^2=8x,\\  y=k(x-2),\end{matrix}\right.$

$可得k^2x^2-(4k^2+8)x+4k^2=0，$

$显然\Delta >0，从而x_1+x_2=\frac{4k^2+8}{k^2}，x_1x_2=4.$

$\therefore k_1+k_2=\frac{y_1}{x_1}+\frac{y_2}{x_2}=\frac{k(x_1-2)}{x_1}+\frac{k(x_2-2)}{x_2}=2k-\left(\frac{2k}{x_1}+\frac{2k}{x_2}\right)=2k-\frac{2k(x_1+x_2)}{x_1x_2}=2k-2k\frac{\frac{4k^2+8}{k^2}}{4}=-\frac{4}{k}=2，\therefore k=-2.$

$\therefore 直线l的方程为2x+y-4=0.$

Final answer: $2x+y-4=0$
Subfield: Conic Sections


Question: $已知抛物线C:y^2=2px(p>0)过点M(1，-2\sqrt{2})，直线l经过抛物线的焦点F与抛物线交于A，B两点.$

![Image](https://cdntk.53zaixian.com/tq/1623894791726759937/img/compression/M_e83aea7715e7c46fe2583d23c2fe92ac.jpg)
$若直线OA，OB的斜率分别为k_1，k_2，且k_1+k_2=2，求直线l的方程.$

Solution:
$由题意知直线l的斜率存在且不为0，$

$设直线l的方程为y=k(x-2)，A(x_1,y_1)，B(x_2,y_2)，$

联立
$\left\{\begin{matrix}y^2=8x,\\  y=k(x-2),\end{matrix}\right.$

$可得k^2x^2-(4k^2+8)x+4k^2=0，$

$显然\Delta >0，从而x_1+x_2=\frac{4k^2+8}{k^2}，x_1x_2=4.$

$\therefore k_1+k_2=\frac{y_1}{x_1}+\frac{y_2}{x_2}=\frac{k(x_1-2)}{x_1}+\frac{k(x_2-2)}{x_2}=2k-\left(\frac{2k}{x_1}+\frac{2k}{x_2}\right)=2k-\frac{2k(x_1+x_2)}{x_1x_2}=2k-2k\frac{\frac{4k^2+8}{k^2}}{4}=-\frac{4}{k}=2，\therefore k=-2.$

$\therefore 直线l的方程为2x+y-4=0.$

Final answer: $2x+y-4=0$
Subfield: Conic Sections


Question: $已知椭圆 C 的方程为 \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a > b > 0)，若右焦点为 F(\sqrt{2}，0)，且离心率为 \frac{\sqrt{6}}{3}。$
求椭圆C的方程；

Solution:
由题意得
$$
\left\{
\begin{matrix}
c=\sqrt{2},\\ 
e=\frac{c}{a}=\frac{\sqrt{6}}{3},\\ 
a^2=b^2+c^2,
\end{matrix}
\right.
$$
解得
$$
\left\{
\begin{matrix}
a^2=3,\\ 
b^2=1,
\end{matrix}
\right.
$$
故椭圆C的方程为
$$
\frac{x^2}{3}+ y^{2}=1.
$$

Final answer: $\frac{x^2}{3}+ y^{2}=1$
Subfield: Conic Sections


Question: $设椭圆 x^2/a^2 + y^2/b^2=1 (a>b>0) 的左焦点为 F，上顶点为 B. 已知椭圆的短轴长为4，离心率为 \sqrt{5}/5。$
求椭圆的方程；

Solution:
$设椭圆的半焦距为c，依题意，2b=4，\frac{c}{a}=\frac{\sqrt{5}}{5}，又a^2=b^2+c^2，可得a=\sqrt{5}，b=2，c=1.所以，椭圆的方程为\frac{x^2}{5}+\frac{y^2}{4}=1.$

思路分析

$根据条件求出基本量a，b得到椭圆方程.$

Final answer: $\frac{x^2}{5}+\frac{y^2}{4}=1$
Subfield: Conic Sections


Question: $已知椭圆E: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)的左，右焦点为F1,F2，离心率e= \frac{2}{3}, P为椭圆E上任意一点，且满足 \overrightarrow{PF_1} \cdot  \overrightarrow{PF_2} 的最小值为1.$
$求椭圆E的标准方程；$

Solution:
设点$P(x_0,y_0), F_1(-c,0), F_2(c,0)，则 \(\frac{x_0^2}{a^2}\) + \(\frac{y_0^2}{b^2}\) = 1，即 \(y_0^2\) = \(b^2(1- \frac{x_0^2}{a^2})\), $$\overrightarrow{PF_1} = (-c - x_0, -y_0), \overrightarrow{PF_2} = (c-x_0,-y_0),$
所以 $\overrightarrow{PF_1} \cdot \overrightarrow{PF_2} = (x_0 + c)(x_0 - c) + y_0^2 = x_0^2 + y_0^2 - c^2 = \frac{a^2 - b^2}{a^2} x_0^2 + b^2 - c^2,$
所以当$x_0=0$时，$\overrightarrow{PF_1} \cdot \overrightarrow{PF_2}取得最小值1，所以 b^2 - c^2 = 1,$
由 $\begin{cases}b^2 - c^2 = 1,\\ e = \frac{c}{a} = \frac{2}{3},\\ a^2 = b^2 + c^2\end{cases} 解得 \begin{cases}a^2 = 9,\\ b^2 = 5,\\ c^2 = 4\end{cases}$
所以椭圆E的标准方程为$\frac{x^2}{9} + \frac{y^2}{5}=1$.

Final answer: $\frac{x^2}{9} + \frac{y^2}{5}=1$
Subfield: Conic Sections


Question: $已知椭圆𝐶：\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(𝑎> 𝑏>0)过点𝐷(-2，0)，且焦距为2\sqrt{3}。$
求椭圆C的方程；

Solution:
$由题意得\left\{\begin{matrix}2c=2\sqrt{3},\\ a=2,\\ a^2=b^2+c^2.\end{matrix}\right.解得\left\{\begin{matrix}a^2=4,\\ b^2=1.\end{matrix}\right.    $
$故椭圆C的方程为\frac{x^2}{4}+y^2=1.$

Final answer: $\frac{x^2}{4}+y^2=1$
Subfield: Conic Sections


Question: $已知椭圆 C：\frac{y^2}{a^2}+\frac{x^2}{b^2}=1(a>b>0)的离心率为\frac{\sqrt{2}}{2}，点 P(1,0)在椭圆 C 上，直线 y=y_0与椭圆 C 交于不同的两点 A，B。$
求椭圆C的方程；

Solution:
由题意得

$$
\left\{\begin{matrix}\frac{1}{b^2}=1,\\ \frac{c}{a}=\frac{\sqrt{2}}{2},\\ a^2=b^2+c^2,\end{matrix}\right.
$$

$解得a^2=2，b^2=1.$
$所以椭圆C的方程为$

$\frac{y^2}{2}$

$+x^2=1.$

Final answer: $\frac{y^2}{2}+x^2=1$
Subfield: Conic Sections


Question: $已知椭圆C: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 (a>b>0)过点(0,\sqrt{3})，且离心率为\frac{1}{2}.设A，B为椭圆C的左、右顶点，P为椭圆上异于A，B的一点，直线AP，BP分别与直线l: x=4相交于M，N两点，且直线MB与椭圆C交于另一点H.$
求椭圆C的标准方程；

Solution:
由
$\left\{\begin{matrix}b=\sqrt{3},\\ \frac{c}{a}=\frac{1}{2},\\ a^2=b^2+c^2,\end{matrix}\right.$
得
$\left\{\begin{matrix}a=2,\\ b=\sqrt{3},\\ c=1,\end{matrix}\right.$
所以椭圆C的标准方程是
$\frac{x^2}{4}+\frac{y^2}{3}=1.$

Final answer: $\frac{x^2}{4}+\frac{y^2}{3}=1$
Subfield: Conic Sections


Question: $已知点P是椭圆C上的任意一点，P到直线l_1：x＝-2的距离为d_1，与点F(-1，0)的距离为d_2，且\frac{d_2}{d_1}=\frac{\sqrt{2}}{2}.$
求椭圆C的方程；

Solution:
$设P(x,y)，则d_1 = |x + 2|，d_2 = \sqrt{{(x+1)}^2 + y^2}，$
$\therefore  \frac{d_2}{d_1} = \frac{\sqrt{{(x+1)}^2 + y^2}}{|x+2|} = \frac{\sqrt{2}}{2}，化简得 \frac{x^2}{2} + y^2 = 1，$
$\therefore 椭圆C的方程为 \frac{x^2}{2} + y^2 = 1.$

Final answer: $\frac{x^2}{2} + y^2 = 1$
Subfield: Conic Sections


Question: $已知椭圆C: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b>0)的离心率是\frac{\sqrt{2}}{2}，且过点P(\sqrt{2},1).直线y=\frac{\sqrt{2}}{2}x+m与椭圆C相交于A,B两点.$
求椭圆C的方程；

Solution:
设椭圆C的半焦距为c.

$由 \frac{c^2}{a^2}= \frac{a^2-b^2}{a^2}=1-\frac{b^2}{a^2}=\frac{1}{2}，得 a^2=2b^2.$

$由 \left\{\begin{matrix}a^2=2b^2,\\ \frac{2}{a^2}+\frac{1}{b^2}=1,\end{matrix}\right. 解得 \left\{\begin{matrix}a^2=4,\\ b^2=2.\end{matrix}\right.$

$所以椭圆C的方程为 \frac{x^2}{4} + \frac{y^2}{2}=1.$

Final answer: $\frac{x^2}{4} + \frac{y^2}{2}=1$
Subfield: Conic Sections


Question: $如图，已知F是抛物线y^2=2px(p＞0)的焦点，M是抛物线的准线与x轴的交点，且|MF|=2.$
<img_2504>
求抛物线的方程；

Solution:
$由题意知p=2，所以抛物线的方程是y^2=4x.$

Final answer: $y^2=4x$
Subfield: Conic Sections


Question: $如图，已知F是抛物线y^2=2px(p＞0)的焦点，M是抛物线的准线与x轴的交点，且|MF|=2.$
![image](https://cdntk.53zaixian.com/tq/1623875912133644289/img/compression/M_5b0d4ffd9338c55a0cb9921c089e4531.jpg)
求抛物线的方程；

Solution:
$由题意知p=2，所以抛物线的方程是y^2=4x.$

Final answer: $y^2=4x$
Subfield: Conic Sections


Question: $已知椭圆C: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a > b > 0) 经过点(2,0)，一个焦点为(\sqrt{3},0)。$
$求椭圆C的方程；$

Solution:
$由题意得a=2，c=\sqrt{3}，因为a^2-b^2=c^2，所以b=1。所以椭圆C的方程是\frac{x^2}{4}+y^2=1。$

Final answer: $\frac{x^2}{4}+y^2=1$
Subfield: Conic Sections


Question: $已知中心在原点，焦点在x轴上的椭圆C过点 (1, \frac{\sqrt{3}}{2})，离心率为 \frac{\sqrt{3}}{2}，点A为其右顶点。过点B(1,0)作直线l与椭圆C相交于E、F两点，直线AE、AF与直线x=3分别交于点M、N。$
$求椭圆 C 的方程；$

Solution:
$设椭圆的标准方程为 x^2/a^2 + y^2/b^2 = 1(a > b > 0)，$
由题意，得 
$$
\begin{aligned}
\left\{
\begin{matrix}
1/a^2 + 3/4b^2=1,\\ 
c/a=\sqrt{3}/2,\\ 
a^2=b^2+c^2,
\end{matrix}
\right.
\end{aligned}
$$
$解得 a^2=4，b^2=1，$
$所以椭圆 C 的标准方程为 x^2/4 + y^2 = 1.$

Final answer: $\frac{x^2}{4} + y^2 = 1$
Subfield: Conic Sections


Question: $已知椭圆 C：\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 (a>b>0)经过点 P(2,1)，P到椭圆 C 的两个焦点的距离和为 4\sqrt{2}.$
求椭圆C的方程；

Solution:
$由题意得 $
$\left\{\begin{matrix}2a=4\sqrt{2},\\ \frac{4}{a^2}+\frac{1}{b^2}=1,\end{matrix}\right.$
$所以a^2=8，b^2=2，$
$所以椭圆C的方程为 \frac{x^2}{8} + \frac{y^2}{2} =1.$

Final answer: $\frac{x^2}{8} + \frac{y^2}{2} =1$
Subfield: Trigonometric Functions


Question: $已知\frac{\pi }{4}<\alpha <\frac{\pi }{2}, f(\alpha )=\frac{2\cos \left(\frac{\pi }{2}+\alpha \right) \cdot \sqrt{1-\sin  2\alpha }}{\tan (\alpha +\pi ) \cdot \sqrt{2+2\cos  2\alpha }}.$
$化简f(\alpha)；$

Solution:
$f(\alpha)=\frac{-2\sin \alpha \cdot \sqrt{\sin ^2\alpha -2\sin \alpha \cdot \cos \alpha +\cos ^2\alpha }}{\tan \alpha \cdot \sqrt{2 \cdot 2\cos ^2\alpha }}$

$=-\frac{\sin \alpha \cdot |\sin \alpha -\cos \alpha |}{\frac{\sin \alpha }{\cos \alpha } \cdot |\cos \alpha |}，\because\frac{\pi }{4}<\alpha<\frac{\pi }{2},$

$\therefore f(\alpha)=-\frac{\sin \alpha \cdot (\sin \alpha -\cos \alpha )}{\frac{\sin \alpha }{\cos \alpha } \cdot \cos \alpha }=\cos \alpha-\sin \alpha.$

Final answer: $\cos \alpha-\sin \alpha$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=Asin(\omega x+\phi )(A>0, \omega >0, |\phi |<\pi )的图象如图所示.$
<img_172>
$求函数f(x)的解析式；$

Solution:
$由题图可知 A=2， \frac{T}{2} =\frac{\pi}{2}, T=\pi , 则 \frac{2\pi}{|\omega|} = \pi , 又 \omega >0, 所以 \omega=2, f(x)=2sin(2x+\phi ). 又函数图象过点 \left(\frac{\pi}{2},-\sqrt{3}\right)，所以 f\left(\frac{\pi}{2}\right) = 2sin(\pi +\phi )=-2sin\phi =-\sqrt{3}, sin\phi =\frac{\sqrt{3}}{2}，由于 |\phi |<\pi ，所以 \phi =\frac{\pi}{3} 或 \phi =\frac{2\pi}{3}。 $

$当 \phi =\frac{\pi}{3} 时， f(x)=2sin\left(2x+\frac{\pi}{3}\right), f\left(\frac{\pi}{12}\right) = 2sin\frac{\pi}{2}=2，结合图象可知不正确。$
$所以 \phi =\frac{2\pi}{3}，则 f(x)=2sin\left(2x+\frac{2\pi}{3}\right).$



Final answer: $f(x)=2sin\left(2x+\frac{2\pi}{3}\right)$
Subfield: Trigonometric Functions


Question: $已知函数f(x)=Asin(\omega x+\phi )(A>0, \omega >0, |\phi |<\pi )的图象如图所示.$
![图象](https://cdntk.53zaixian.com/tq/1623608893467144195/img/compression/M_ebc0957cc06dd2bc791e92939320419a.jpg)
$求函数f(x)的解析式；$

Solution:
$由题图可知 A=2， \frac{T}{2} =\frac{\pi}{2}, T=\pi , 则 \frac{2\pi}{|\omega|} = \pi , 又 \omega >0, 所以 \omega=2, f(x)=2sin(2x+\phi ). 又函数图象过点 \left(\frac{\pi}{2},-\sqrt{3}\right)，所以 f\left(\frac{\pi}{2}\right) = 2sin(\pi +\phi )=-2sin\phi =-\sqrt{3}, sin\phi =\frac{\sqrt{3}}{2}，由于 |\phi |<\pi ，所以 \phi =\frac{\pi}{3} 或 \phi =\frac{2\pi}{3}。 $

$当 \phi =\frac{\pi}{3} 时， f(x)=2sin\left(2x+\frac{\pi}{3}\right), f\left(\frac{\pi}{12}\right) = 2sin\frac{\pi}{2}=2，结合图象可知不正确。$
$所以 \phi =\frac{2\pi}{3}，则 f(x)=2sin\left(2x+\frac{2\pi}{3}\right).$

Final answer: $f(x)=2sin\left(2x+\frac{2\pi}{3}\right)$
Subfield: Plane Geometry


Question: $如图，平面四边形OABC中，OA=OB=OC=1，对角线AC,OB相交于点M．设 \overrightarrow{AM}=\lambda \overrightarrow{AC}(0<\lambda <1)，且 \overrightarrow{OM}=t\overrightarrow{OB}(0<t<1).$

<img_145>
$用向量\overrightarrow{OA},\overrightarrow{OB}表示向量\overrightarrow{OC}；$

Solution:
$因为 \overrightarrow{AM}=\lambda\overrightarrow{AC}(0<\lambda<1)，\overrightarrow{OM}=t\overrightarrow{OB}(0<t<1)，所以有 \overrightarrow{OA}=\overrightarrow{OM}+\overrightarrow{MA}=t\overrightarrow{OB}-\lambda\overrightarrow{AC}=t\overrightarrow{OB}-\lambda(\overrightarrow{OC}-\overrightarrow{OA})，即\lambda\overrightarrow{OC}=(\lambda-1)\overrightarrow{OA}+t\overrightarrow{OB}，所以\overrightarrow{OC}=\frac{\lambda -1}{\lambda }\overrightarrow{OA}+\frac{t}{\lambda }\overrightarrow{OB}.$

Final answer: $\overrightarrow{OC}=\frac{\lambda -1}{\lambda }\overrightarrow{OA}+\frac{t}{\lambda }\overrightarrow{OB}$
Subfield: Plane Geometry


Question: $如图，平面四边形OABC中，OA=OB=OC=1，对角线AC,OB相交于点M．设 \overrightarrow{AM}=\lambda \overrightarrow{AC}(0<\lambda <1)，且 \overrightarrow{OM}=t\overrightarrow{OB}(0<t<1).$

![image](https://cdntk.53zaixian.com/tq/1623609363186130946/img/compression/M_3aa749f047c9d316d5976dfdaabff2f6.jpg)
$用向量\overrightarrow{OA},\overrightarrow{OB}表示向量\overrightarrow{OC}；$

Solution:
$因为 \overrightarrow{AM}=\lambda\overrightarrow{AC}(0<\lambda<1)，\overrightarrow{OM}=t\overrightarrow{OB}(0<t<1)，所以有 \overrightarrow{OA}=\overrightarrow{OM}+\overrightarrow{MA}=t\overrightarrow{OB}-\lambda\overrightarrow{AC}=t\overrightarrow{OB}-\lambda(\overrightarrow{OC}-\overrightarrow{OA})，即\lambda\overrightarrow{OC}=(\lambda-1)\overrightarrow{OA}+t\overrightarrow{OB}，所以\overrightarrow{OC}=\frac{\lambda -1}{\lambda }\overrightarrow{OA}+\frac{t}{\lambda }\overrightarrow{OB}.$

Final answer: $\overrightarrow{OC}=\frac{\lambda -1}{\lambda }\overrightarrow{OA}+\frac{t}{\lambda }\overrightarrow{OB}$
Subfield: Plane Geometry


Question: $如图，平面四边形OABC中，OA=OB=OC=1，对角线AC,OB相交于点M．设 \overrightarrow{AM}=\lambda \overrightarrow{AC}(0<\lambda <1)，且 \overrightarrow{OM}=t\overrightarrow{OB}(0<t<1).$

<img_145>
$若\angle BOA=\frac{\pi }{3}，记\lambda =f(t)，求f(t)的解析式．$

Solution:
$\because \angle BOA=\frac{\pi }{3}，OA=OB=1，\therefore \overrightarrow{OA}\cdot \overrightarrow{OB}=|\overrightarrow{OA}|\cdot |\overrightarrow{OB}|\cdot cos \frac{\pi }{3}=\frac{1}{2}，又\overrightarrow{OC}=\frac{\lambda -1}{\lambda } \overrightarrow{OA}+\frac{t}{\lambda } \overrightarrow{OB}且OC=1，\therefore \overrightarrow{OC}^2=\left(\frac{\lambda -1}{\lambda }\overrightarrow{OA}+\frac{t}{\lambda }\overrightarrow{OB}\right)^{2} =1，即\left(\frac{\lambda -1}{\lambda }\right)^{2} +\left(\frac{t}{\lambda }\right)^{2} +\frac{\lambda -1}{\lambda }\cdot \frac{t}{\lambda }=1，\therefore \lambda ^2-2\lambda +1+t^2+\lambda t-t=\lambda ^2，\therefore \lambda =f(t)=\frac{t^2-t+1}{2-t}(0<t<1).$

Final answer: $f(t)=\frac{t^2-t+1}{2-t}$
Subfield: Plane Geometry


Question: $如图，平面四边形OABC中，OA=OB=OC=1，对角线AC,OB相交于点M．设 \overrightarrow{AM}=\lambda \overrightarrow{AC}(0<\lambda <1)，且 \overrightarrow{OM}=t\overrightarrow{OB}(0<t<1).$

![image](https://cdntk.53zaixian.com/tq/1623609363186130946/img/compression/M_3aa749f047c9d316d5976dfdaabff2f6.jpg)
$若\angle BOA=\frac{\pi }{3}，记\lambda =f(t)，求f(t)的解析式．$

Solution:
$\because \angle BOA=\frac{\pi }{3}，OA=OB=1，\therefore \overrightarrow{OA}\cdot \overrightarrow{OB}=|\overrightarrow{OA}|\cdot |\overrightarrow{OB}|\cdot cos \frac{\pi }{3}=\frac{1}{2}，又\overrightarrow{OC}=\frac{\lambda -1}{\lambda } \overrightarrow{OA}+\frac{t}{\lambda } \overrightarrow{OB}且OC=1，\therefore \overrightarrow{OC}^2=\left(\frac{\lambda -1}{\lambda }\overrightarrow{OA}+\frac{t}{\lambda }\overrightarrow{OB}\right)^{2} =1，即\left(\frac{\lambda -1}{\lambda }\right)^{2} +\left(\frac{t}{\lambda }\right)^{2} +\frac{\lambda -1}{\lambda }\cdot \frac{t}{\lambda }=1，\therefore \lambda ^2-2\lambda +1+t^2+\lambda t-t=\lambda ^2，\therefore \lambda =f(t)=\frac{t^2-t+1}{2-t}(0<t<1).$

Final answer: $f(t)=\frac{t^2-t+1}{2-t}$
Subfield: Sequence


Question: $记各项均为正数的数列a_n的前n项和是S_n，已知a_n^2 + a_n = 2S_n，n为正整数.$
$求a_n的通项公式；$

Solution:
$由a_n^2+a_n=2S_n得当n\geq 2时，a_{n-1}^2+a_{n-1}=2S_{n-1},两式相减得a_n^2-a_{n-1}^2+a_n-a_{n-1}=2a_n，$
$即(a_n-a_{n-1}-1)(a_n+a_{n-1})=0，因为数列各项均为正数，所以a_n=a_{n-1}+1(n\geq 2)，即a_n-a_{n-1}=1(n\geq 2),$
$故{a_n}是公差为1的等差数列，又当n=1时，a_1^2+a_1=2a_1,解得a_1=1,所以a_n=n.$

Final answer: $a_n=n$
Subfield: Sequence


Question: $记各项均为正数的数列a_n的前n项和是S_n，已知a_n^2 + a_n = 2S_n，n为正整数.$
$设b_n=tan a_n\cdot tan a_{n+1}，求数列{b_n}的前n项和T_n.$

Solution:
$\\tan  1=\\tan [(n+1)-n]=\frac{\\tan  (n+1)-\\tan   n}{1+\\tan  (n+1)\\tan   n}$

$故 b_n=\tan(n+1)\tan n=\frac{\tan (n+1)-\tan  n}{\tan  1}-1$

$则 T_n=b_1+b_2+b_3⋯+b_n=\frac{1}{\tan  1}[\tan(n+1)-\tan n+\tan n-\tan(n-1)+⋯+\tan 2-\tan 1]-n=\frac{1}{\tan  1}[\tan(n+1)-\tan 1]-n=\frac{\tan (n+1)}{\tan  1}-n-1$

Final answer: $\frac{\tan (n+1)}{\tan  1}-n-1$
Subfield: Sequence


Question: $已知数列a_n的前n项和为S_n，a_1=-11，a_2=-9，且S_{n+1}+S_{n-1}=2S_n+2 (n\geq 2)．$
$求数列{a_n}的通项公式；$

Solution:
：
$由题意知(S_{n+1}-S_n)-(S_n-S_{n-1})=2(n\geq 2)，则a_{n+1}-a_n=2(n\geq 2)，$
$又a_2-a_1=2，所以{a_n}是首项为-11，公差为2的等差数列，则a_n=a_1+(n-1)d=2n-13.$

Final answer: ${a_n}=2n-13$
Subfield: Sequence


Question: $已知数列{a_n}的前n项和S_n满足S_n=2a_n-1.$
$求数列a_n的通项公式；$

Solution:
$当n=1时，S_1=2a_1-1，解得a_1=1。$

$当n\geq 2时，S_{n-1}=2a_{n-1}-1，则S_n-S_{n-1}=a_n=2a_n-2a_{n-1}，即a_n=2a_{n-1}。$

$所以{a_n}是以1为首项，2为公比的等比数列，所以a_n=2^{n-1} (n\in N^)。$

Final answer: a_n=2^{n-1}
Subfield: Sequence


Question: $已知数列{a_n}的前n项和S_n满足S_n=2a_n-1.$
$记b_n = \frac{2n-1}{a_n}，求数列{b_n}的前n项和T_n.$

Solution:
$b_n=\frac{2n-1}{a_n}=\frac{2n-1}{2^{n-1}}，\therefore T_n=1+\frac{3}{2^1}+\frac{5}{2^2}+\ldots +\frac{2n-1}{2^{n-1}}$

$\therefore \frac{1}{2}T_n=\frac{1}{2^1}+\frac{3}{2^2}+\frac{5}{2^3}+\ldots +\frac{2n-3}{2^{n-1}}+\frac{2n-1}{2^n}$

$两式相减得 \frac{1}{2}T_n=1+\frac{2}{2^1}+\frac{2}{2^2}+\frac{2}{2^3}+\ldots +\frac{2}{2^{n-1}}-\frac{2n-1}{2^n}=3-\frac{2n+3}{2^n}，\therefore T_n=6-\frac{2n+3}{2^{n-1}}(n \in N^).$

Final answer: $T_n=6-\frac{2n+3}{2^{n-1}}$
Subfield: Sequence


Question: $已知数列{a_n}满足a_na_{n+1}=2^{2n},a_1=1.$
$求a_{2n}；$

Solution:
$因为a_n a_{n+1} = 2^{2n}, a_1 = 1, 所以a_1a_2 = 2^2, a_2 = 4, 又a_{n+1}a_{n+2} = 2^{2n+2}，所以\frac{a_{n+1}a_{n+2}}{a_na_{n+1}} = \frac{a_{n+2}}{a_n} = \frac{2^{2n+2}}{2^{2n}} = 4，所以{a_n}的奇数项是以1为首项，4为公比的等比数列，偶数项是以4为首项，4为公比的等比数列，所以a_n = \begin{cases} 2^{n-1}, & \text{if } n \text{ is odd} \\ 2^n, & \text{if } n \text{ is even} \end{cases}, 所以a_{2n} = 2^{2n}.$

Final answer: $2^{2n}$
Subfield: Sequence


Question: $已知正项数列{a_n}，其前n项和S_n满足a_n(2S_n-a_n)=1(n\in N^).$
$求证：数列{S^2_n}是等差数列，并求出S_{n}的表达式；$

Solution:
$依题意，得正项数列a_n中，a^2_1=1，即a_1=1，当n\geq 2时，a_n=S_n-S_{n-1}，即(S_n-S_{n-1})[2S_n-(S_n-S_{n-1})]=1，整理得S^2_n-S^2_{n-1}=1，又S^2_1=a^2_1=1，因此，数列S^2_n是以1为首项，1为公差的等差数列，则S^2_n=n，因为a_n是正项数列，即S_n>0，所以S_n=\sqrt{n}.$

Final answer: $S_n=\sqrt{n}$
Subfield: Sequence


Question: $等差数列{a_n}的前n项和为S_n，已知a_{10}=30，a_{20}=50.$
$求通项公式a_n；$

Solution:
$设等差数列a_n的公差为d，$
依题意有
$$
\begin{align*}
a_{10}&=a_1+9d=30,\\ 
a_{20}&=a_1+19d=50,
\end{align*}
$$
解得
$$
\begin{align*}
a_1&=12,\\ 
d&=2,
\end{align*}
$$
$所以a_n=2n+10 (n\in N^).$

Final answer: $a_n=2n+10$
Subfield: Sequence


Question: $已知数列a_n满足a_1=2，a_{n+1}=2a_n-n+1 (n\in N^).$
$证明：数列{a_n - n}是等比数列，并求出数列{a_n}的通项公式；$

Solution:
$因为数列{a_n}满足a_1=2，a_{n+1}=2a_{n}-n+1，所以a_{n+1}-(n+1)=2(a_{n}-n)，而a_1-1=1，于是数列{a_n-n}是首项为1，公比为2的等比数列，a_n-n=1\cdot 2^{n-1}，即a_{n}=2^{n-1}+n.$
Subfield: Sequence


Question: $已知数列a_n满足a_1=2，a_{n+1}=2a_n-n+1 (n\in N^).$
$数列b_n满足：b_n=\frac{n}{2a_n-2n}(n\in N^{})，求数列b_n的前n项和S_n.$

Solution:
$由(1)知，b_n=\frac{n}{2(2^{n-1}+n)-2n}=\frac{n}{2^n}，S_n=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots +\frac{n}{2^n}，$

$则\frac{1}{2}S_n=\frac{1}{2^2}+\frac{2}{2^3}+\frac{3}{2^4}+\ldots +\frac{n-1}{2^n}+\frac{n}{2^{n+1}}，$

$于是得\frac{1}{2}S_n=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\ldots +\frac{1}{2^n}-\frac{n}{2^{n+1}}=\frac{\frac{1}{2}\left[1-\left(\frac{1}{2}\right)^n\right]}{1-\frac{1}{2}}-\frac{n}{2^{n+1}}=1-\frac{1}{2^n}-\frac{n}{2^{n+1}}=1-\frac{n+2}{2^{n+1}}，$

$S_n=2-\frac{n+2}{2^n}，$

$所以数列{b_n}的前n项和S_n=2-\frac{n+2}{2^n}。$

Final answer: $S_n=2-\frac{n+2}{2^n}$
Subfield: Sequence


Question: $设{a_n}是等差数列，a_1=-10，且a_2+10，a_3+8,a_4+6成等比数列.$
$求a_n的通项公式；$

Solution:
$设a_n的公差为d.因为a_1=-10，$
$所以a_2=-10+d,a_3=-10+2d,a_4=-10+3d.$
$因为a_2+10,a_3+8,a_4+6成等比数列，$
$所以(a_3+8)^2=(a_2+10)(a_4+6).$
$所以(-2+2d)^2=d(-4+3d).解得d_1=d_2=2.$
$所以a_n=a_1+(n-1)d=2n-12.$

Final answer: $a_n=2n-12$
Subfield: Sequence


Question: $设等差数列a_n的公差为d,且d>1,令b_n=\frac{n^2+n}{a_n},记S_n,T_n分别为数列a_n,b_n的前n项和.$
$若3a^2=3a^1+a^3, S^3+T^3=21,求{a^n}的通项公式；$

Solution:
：

$因为 3a_2 = 3a_1 + a_3，所以 3(a_1 + d) = 3a_1 + a_1 + 2d，$

$因此 a_1 = d > 1，所以 S_3 = a_1 + a_2 + a_3 = a_1 + a_1 + d + a_1 + 2d = 6a_1，$

$又因为 b_n = \frac{n^2+n}{a_n}，所以 b_1 = \frac{2}{a_1}，b_2 = \frac{6}{a_2} = \frac{6}{a_1+d} = \frac{3}{a_1}，b_3 = \frac{12}{a_3} = \frac{12}{a_1+2d} = \frac{4}{a_1}，$

$所以 T_3 = b_1 + b_2 + b_3 = \frac{9}{a_1}，$

$因此 S_3 + T_3 = 6a_1 + \frac{9}{a_1} = 21，$

$解得 a_1 = 3 或 a_1 = \frac{1}{2}(舍弃)，所以 a_n = 3n。$

Final answer: $a_n = 3n$
Subfield: Sequence


Question: $记S_n为数列{a_n}的前n项和，b_n为数列{S_n}的前n项积，已知 \frac{2}{S_n} + \frac{1}{b_n} = 2.$
$求a_n的通项公式.$

Solution:
$第一步：写出等差数列的通项公式 b_n，并利用 S_n 与 b_n 的关系写出 S_n .$
$由(1)知，b_n = \frac{3}{2} + (n - 1) \times  \frac{1}{2} = \frac{n+2}{2}，$
$故当 n\geq 2 时，S_n = \frac{b_n}{b_{n-1}} = \frac{n+2}{n+1}，S_1 也符合该式，$
$即 S_n = \frac{n+2}{n+1}， n\in N^。$

$第二步：利用前 n 项和 S_n 求通项 a_n 。$
$当 n = 1 时，a_1 = S_1 = \frac{3}{2}，$
$当 n \geq  2 时，a_n = S_n - S_{n-1} = \frac{n+2}{n+1} - \frac{n+1}{n}= -\frac{1}{n(n+1)}, a_1 不符合该式，$
$所以 a_n = \begin{cases} \frac{3}{2}, & n=1,\\ -\frac{1}{n(n+1)}, & n \geq 2. \end{cases}$

Final answer: $a_n = \begin{cases} \frac{3}{2}, & n=1,\\ -\frac{1}{n(n+1)}, & n \geq 2. \end{cases}$
Subfield: Sequence


Question: $记S_n为等差数列{a_n}的前n项和，已知a_2=11，S_{10}=40.$
$求a_n的通项公式；$

Solution:
$设等差数列a_n的公差为d$

则有以下等式：

$\begin{cases}
a_1+d=11, \\
10a_1+45d=40, \\
\end{cases}$

解得：

$\begin{cases}
a_1=13, \\
d=-2, \\
\end{cases}$

$\therefore  a_n=13-2(n-1)=15-2n.$

Final answer: $a_n=15-2n$
Subfield: Sequence


Question: $记S_n为等差数列{a_n}的前n项和，已知a_2=11，S_{10}=40.$
$求数列{|a_n|}的前n项和T_n.$

Solution:
$由a_n=15-2n知，当n\leq 7, n\in N时，a_n>0，当n\geq 8, n\in N时，a_n<0，$
$\therefore 当n\leq 7, n\in N时，T_n=|a_1|+|a_2|+\ldots +|a_n|=a_1+a_2+\ldots +a_n=S_n=\frac{n(28-2n)}{2}=-n^2+14n,$
$当n\geq 8, n\in N时，T_n=(a_1+a_2+\ldots +a_7)-(a_8+a_9+\ldots +a_n)$
$=2S_7-S_n=98-(-n^2+14n)=n^2-14n+98.$
$\therefore T_n=\left\{\begin{matrix}-n^2+14n(n\leq 7{\text{且}}n\in N),\\ n^2-14n+98(n\geq 8{\text{且}}n\in N).\end{matrix}\right.$

Final answer: $T_n=\left\{\begin{matrix}-n^2+14n(n\leq 7{\text{且}}n\in N),\\ n^2-14n+98(n\geq 8{\text{且}}n\in N).\end{matrix}\right.$
Subfield: Sequence


Question: $记S_n为等差数列{a_n}的前n项和.已知S_9=-a_5.$
$若a_3=4,求{a_n}的通项公式；$

Solution:
$设a_n的公差为d.$

$由S_9=-a_5得$

$\frac{9(a_1+a_9)}{2}$

$=-a_5, 化简得a_5=0,即a_1+4d=0.$

$由a_3=4得 a_1+2d=4.$

$于是a_1=8,d=-2.$

$因此a_n的通项公式为a_n=10-2n.$

Final answer: ${a_n}=10-2n$
Subfield: Sequence


Question: $记S_n为公差不为零的等差数列{a_n}的前n项和，若a_3=S_5,a_2\cdot a_4=S_4.$
$求a_n的通项公式；$

Solution:
解法一：
$设等差数列a_n的公差为d(d\neq 0)，则a_3=S_5\Rightarrow a_1+2d=5a_1+10d\Rightarrow 4a_1+8d=0\Rightarrow a_1+2d=0\Rightarrow a_1=-2d，①$
$a_2\cdot a_4=S_4\Rightarrow (a_1+d)(a_1+3d)=4a_1+6d，②  $
$将①代入②得-d^2=-2d\Rightarrow d=0(舍)或d=2，$
$\therefore a_1=-2d=-4,\therefore a_n=-4+(n-1)\times 2=2n-6.$

解法二：
$由等差数列的性质可得S_5=5a_3，则a_3=5a_3,\therefore a_3=0，$
$设等差数列的公差为d，$
$从而有a_2.a_4=(a_3-d)(a_3+d)=-d^2，$
$S_4=a_1+a_2+a_3+a_4=(a_3-2d)+(a_3-d)+a_3+(a_3+d)=-2d，$
$从而-d^2=-2d，由于公差不为零，故d=2，$
$所以数列{a_n}的通项公式为a_n=a_3+(n-3)d=2n-6.$

Final answer: $a_n=2n-6$
Subfield: Sequence


Question: $设a_n是等差数列，且a_1=ln 2，a_2+a_3=5ln 2.$
$求a_n的通项公式；$

Solution:
$设a_n的公差为d.$

$因为a_2+a_3=5ln 2,所以2a_1+3d=5ln 2.$

$又a_1=ln 2,所以d=ln 2.所以a_n=a_1+(n-1)d=nln 2.$

Final answer: a_n=n\ln 2
Subfield: Sequence


Question: $设a_n是等差数列，且a_1=ln 2，a_2+a_3=5ln 2.$
$求e^{a_1}+e^{a_2}+\ldots +e^{a_n}.$

Solution:
$因为 e^{a_1} = e^{ln 2} = 2, \frac{e^{a_n}}{e^{a_{n-1}}} = e^{a_n - a_{n-1}} = e^{ln 2} = 2$
$所以{e^{a_n}}是首项为2，公比为2的等比数列.$
$所以 e^{a_1} + e^{a_2} + \ldots  + e^{a_n} = 2 \times \frac{1 - 2^n}{1 - 2} = 2^{n + 1} - 2.$

Final answer: $2^{n + 1} - 2$
Subfield: Sequence


Question: $已知等差数列{a_n}和等比数列{b_n}满足a_1=b_1=1, a_2+a_4=10, b_2 b_4=a_5.$
$求a_n的通项公式；$

Solution:
$设等差数列{a_n}的公差为d.因为a_2+a_4=10，所以2a_1+4d=10，又a_1=1，所以d=2。所以a_n=2n-1.$

Final answer: a_n=2n-1
Subfield: Sequence


Question: $已知等差数列{a_n}和等比数列{b_n}满足a_1=b_1=1, a_2+a_4=10, b_2 b_4=a_5.$
$求和：b_1+b_3+b_5+...+b_{2n-1}.$

Solution:
$设等比数列b_n的公比为q.因为b_2b_4=a_5,所以b_1qb_1q^3=9. 又b_1=1,所以q^2=3. 所以b_{2n-1}=b_1q^{2n-2}=3^{n-1}. 从而b_1+b_3+b_5+\ldots +b_{2n-1}=1+3+3^2+\ldots +3^{n-1} = \frac{3^n-1}{2}.$

Final answer: $\frac{3^n-1}{2}$
Subfield: Sequence


Question: $设等比数列{a_n}满足a_1+a_2=4，a_3-a_1=8.$
$求a_n的通项公式;$

Solution:
$设a_n的公比为q，则a_n=a_1q^{n-1}。$
由已知得

$\left\{\begin{matrix}a_1+a_1q=4,\\ a_1q^2-a_1=8.\end{matrix}\right.$

$解得a_1=1，q=3。$
$所以a_n的通项公式为a_n=3^{n-1}。$

Final answer: $a_n=3^{n-1}$
Subfield: Sequence


Question: $记S_n为数列{a_n}的前n项和，已知a_2=1, 2S_n=na_n。$
$求a_n的通项公式；$

Solution:
$当n=1时，2a_{1}=a_{1},即a_{1}=0,$

$当n\geq 2时，2S_{n-1}=(n-1)a_{n-1},①$

$又2S_{n}=na_{n}，②$

$\therefore ②-①得2a_{n}=na_{n}-(n-1)a_{n-1},$

$即(n-2)a_{n}=(n-1)a_{n-1}.$

$当n=2时，上式成立.$

$当n\geq 3时，\frac{a_n}{a_{n-1}}=\frac{n-1}{n-2},\therefore a_{n}=\frac{a_3}{a_2}\cdot \frac{a_4}{a_3}\cdot \frac{a_5}{a_4}\cdot \ldots \cdot \frac{a_n}{a_{n-1}}\cdot a_{2}=\frac{2}{1}\times \frac{3}{2}\times \frac{4}{3}\cdot \ldots \cdot \frac{n-1}{n-2}\cdot 1=n-1,即a_{n}=n-1(n\geq 3).$

$当n=1时，a_{1}=0符合上式，当n=2时，a_{2}=1符合上式.$

$综上，{a_{n}}的通项公式为a_{n}=n-1,n\in N^{}.$

Final answer: $a_{n}=n-1$