数列-清华北大金秋营试题

(2019 AMC 12B) Define a sequence recursively by x0=5 andxn+1=xn2+5xn+4xn+6for all nonnegative integers n. Let m be the least positive integer such thatxm4+1220.In which of the following intervals does m lie?

(A) [9,26](B) [27,80](C) [81,242](D) [243,728](E) [729,)

 

(2019 AMC 12A) A sequence of numbers is defined recursively by a1=1, a2=37, andan=an2an12an2an1for all n3 Then a2019 can be written as pq, where p and q are relatively prime positive integers. What is p+q?

(A) 2020(B) 4039(C) 6057(D) 6061(E) 8078

 

 

\section{清华大学数学系2020年“大中衔接”研讨与教学活动}

\begin{center} \textbf{2020年清华大学“大中衔接”试题} \end{center}


1.设指数α1,α2,,αn两两不同,系数a1,a2,,an不全为零.证明:
f(x)=a1xα1+a2xα2++anxαn
(0,+)上至多有n1个零点.

2.设f:[1,1]R是连续函数,证明:
limλ[11|λ|λ2+x2f(x)dx]=πf(0).

3.设整数n>1,证明:至多只有有限多个正整数a,使得方程
x12+x22++xn2=ax1x2xn
有非零整数解.

%\let\oldwidering\widering
%\let\widering\undefined
%\usepackage{yhmath} %弧AB
%\let\widering\oldwidering

4.设ABC是单位球面S上三个不同点且都位于第一象限中,对于任何两个不同点P,QS,只要PQ不是S的直径,则平面OPQS的交集是S上的一个圆周, PQ将此圆分成两段弧,将其中较短的那段弧记为\wideparenPQ.设\wideparenBC,\wideparenCA,\wideparenAB的中点分别为DEF.证明: \wideparenAD,\wideparenBE,\wideparenCF经过同一个点.

5.设共有k4个不同的字母,可用它们构成单词.给定一族(记为T)禁用单词,其中任何两个禁用单词长度不等.称一个单词是“可用的”,如果它不含连续一段字母恰为某禁用单词.证明:至少有(k+k24k2)n个长为n的可用单词.

6.设n,p是正整数,集合A1,A2,,Ak{1,2,,n}的子集,满足对任何ij都有|AiAj|p.证明:
k(p1)!n!(n+p12)!(n+p12)!.

\section{2020年北京大学金秋营试题}

\begin{center}
\textbf{2020年北京大学金秋营}

\textbf{第一天}
\end{center}

1.对于非负实数a1,a2,,an,考虑如下2n个实数
i1a1+i2a2++inan,其中ik=±1,1kn,记S为这2n个数中所有正数之和,在i=1nan=1的条件下,求S的最小值.


2.在ABC中, DBC, E,F分别为\wideparenBC,\wideparenBAC, 取ABC外接圆ω, ADE外接圆与射线AB,AC交于点J,K, ADF外接圆与射线AB,AC交于点L,M,证明:若ADJKLM共点,则JMLK交点在ω上.

3.数列{an}满足: a0=2,a1=5,an+1=5anan1,n1,已知ama2n2+a2n1,求证: 3m,3n.

4.求k的最小值,使得将7×7方格挖去k个格后,剩余图形不存在T字形. (T字形指一个方格与其相邻的三个方格有公共边构成的图形)

\begin{center} \textbf{第二天} \end{center}

5. ABC内部取一点O,直线AO,BO,CO分别交对边于D,E,F,若四边形AEFO,BDFO,CDEO都有内切圆,求证: OABC内心和垂心所在直线上

6.若自然数n可以写成若干个自己的不同的因数的和,其中有个为1,就称n为好数,证明:对任意m大于1,存在无穷个m的正倍数为好数,且最小的倍数不大于pm,其中pm最大的奇素因数(若m为二的幂,则p3)

7.求证: r=0pt=1p(1)t1CtrCptCp(pt)p(kr)Cp2kp(modpp) ,其中p为素奇数, k[1,p].

8.求所有的n,使得平面上有n个完全相同的凸多边形,且满足对任意k个凸多边形,所有在它们之中且不在其余多边形中的点的集合为凸多边形(非退化).

 

(Austrian Regional Competition For Advanced Students 2019) Let x,y be real numbers such that (x+1)(y+2)=8. Prove that(xy10)264.

16=(2x+2)(y+2)(2x+2+y+2)24=(10xy)24

(Austrian Regional Competition For Advanced Students 2018) If a,b are positive reals such that a+b<2. Prove that11+a2+11+b221+aband determine all a,b yielding equality.

Proposed by Gottfried Perz

Because21+ab11+a211+b2=(ab)2(1ab)(1+ab)(1+a2)(1+b2)0.

(Austria-Poland 2004 system of equations) Solve the following system of equations in R where all square roots are non-negative:
$$
\begin{matrix}

a - \sqrt{1-b^2} + \sqrt{1-c^2} = d \\
b - \sqrt{1-c^2} + \sqrt{1-d^2} = a \\
c - \sqrt{1-d^2} + \sqrt{1-a^2} = b \\
d - \sqrt{1-a^2} + \sqrt{1-b^2} = c \\

\end{matrix}
$$

Apply the substitution a=sinα, ..., where α,...[π/2,π/2].
Summing up (1) and (2) we obtain sinβcosβ=sinδcosδ. It follows that β=δ OR β+δ=π/2. The same for α and γ. Now it is routine to analyze 3 variants.


(2019 AMC 12B) Define a sequence recursively by x0=5 andxn+1=xn2+5xn+4xn+6for all nonnegative integers n. Let m be the least positive integer such thatxm4+1220.In which of the following intervals does m lie?

(A) [9,26](B) [27,80](C) [81,242](D) [243,728](E) [729,)

 

(2019 AMC 12A) A sequence of numbers is defined recursively by a1=1, a2=37, andan=an2an12an2an1for all n3 Then a2019 can be written as pq, where p and q are relatively prime positive integers. What is p+q?

(A) 2020(B) 4039(C) 6057(D) 6061(E) 8078


(Austria-Poland 1997 Problem) Numbers 491,492,...,4997 are writen on a blackboard. Each time, we can replace two numbers (like a,b) with 2abab+1. After 96 times doing that prenominate action, one number will be left on the board. Find all the possible values fot that number.

Cute; nice to find the invariant Π=Πn=i=1n(2ai1), where n is the number of the numbers written on the blackboard! Since Π97=1, it follows the last number will also be Π1=1.
(This is obviously so, because 2(2abab+1)1=(2a1)(2b1)).

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