《2025年暑假生活北京师范大学出版社高二数学人教版》

1. 已知数列$\sqrt {2},\sqrt {5},2\sqrt {2},\sqrt {11},…$,则$2\sqrt {5}$是这个数列的(B).
A.第6项
B.第7项
C.第8项
D.第9项

答案： B 解析:由数列前几项可知通项公式为$a_{n}= \sqrt{3n-1}$,所以$\sqrt{3n-1}=2\sqrt{5}$,所以$n=7$,所以$2\sqrt{5}$是这个数列的第7项,故选B.

2. 已知数列$\{ a_{n}\}对任意的p,q∈N^{*}满足a_{p+q}= a_{p}+a_{q}$,且$a_{2}= -6$,那么$a_{10}$等于(C).
A.-165
B.-33
C.-30
D.-21

答案： C 解析:因为对任意的$p,q\in \mathbf{N}^{*}$满足$a_{p+q}=a_{p}+a_{q}$,所以$p=q=n$时,有$a_{2n}=2a_{n}$.又因为$a_{2}=-6$,所以$a_{8}=2a_{4}=4a_{2}=-24$,故$a_{10}=a_{2}+a_{8}=-30$.

3. 在等差数列$\{ a_{n}\}$中,若$a_{3}+a_{4}+a_{5}= 12$,则$a_{1}+a_{2}+… +a_{7}$等于(C).
A.14
B.21
C.28
D.35

答案： C 解析:由$a_{3}+a_{4}+a_{5}=12$,得$3a_{4}=12$,所以$a_{4}=4$,所以$a_{1}+a_{2}+\cdots +a_{7}=7a_{4}=28$.

4. 若$\{ a_{n}\}$是等差数列,则下列数列中仍为等差数列的有(C).
①$\{ |a_{n}|\}$;②$\{ a_{n+1}-a_{n}\}$;③$\{ pa_{n}+q\}$(p,q为常数);④$\{ 2a_{n}+n\}$.

A.1个
B.2个
C.3个
D.4个

答案： C 解析:数列$-1,1,3$是等差数列,取绝对值后为1,1,3,不是等差数列,①不符合;若$\{a_{n}\}$是等差数列,利用等差数列的定义,知$\{a_{n+1}-a_{n}\}$为常数列,故是公差为0的等差数列,②符合;设数列$\{a_{n}\}$的公差为$d$,则$pa_{n}+q-(pa_{n-1}+q)=p(a_{n}-a_{n-1})=pd(n\geq 2)$为常数,故$\{pa_{n}+q\}$是等差数列,③符合;设数列$\{a_{n}\}$的公差为$d$,则$2a_{n}+n-(2a_{n-1}+n-1)=2(a_{n}-a_{n-1})+1=2d+1(n\geq 2)$为常数,故$\{2a_{n}+n\}$是等差数列,④符合.故选C.

5. (多选)已知无穷等差数列$\{ a_{n}\}$的前n项和为$S_{n},S_{6}\lt S_{7}$,且$S_{7}＞S_{8}$,则(AD).
A.在数列$\{ a_{n}\}$中,$a_{1}$最大
B.在数列$\{ a_{n}\}$中,$a_{3}或a_{4}$最大
C.$S_{3}= S_{10}$
D.当$n≥8$时,$a_{n}＜0$

答案： AD 解析:因为$S_{6}\lt S_{7},S_{7}\gt S_{8}$,所以$S_{7}-S_{6}=a_{7}\gt 0$,$S_{8}-S_{7}=a_{8}\lt 0$,所以数列$\{a_{n}\}$是递减的等差数列,最大项为$a_{1}$,所以A正确,B错误,D正确;$S_{10}-S_{3}=a_{4}+a_{5}+\cdots +a_{10}=7a_{7}\gt 0$,即$S_{10}\gt S_{3}$,故C错误.

6. 已知数列$\{ a_{n}\}满足2a_{1}+2^{2}a_{2}+2^{3}a_{3}+… +2^{n}a_{n}= n(n∈N^{*})$,若$b_{n}= \frac {1}{log_{2}a_{n}\cdot log_{2}a_{n+1}}$,则数列$\{ b_{n}\}$的前2 021项和$S_{2021}$等于(C).
A.$\frac {2021}{2020}$
B.$\frac {2020}{2021}$
C.$\frac {2021}{2022}$
D.$\frac {2022}{2021}$

答案： C 解析:由$2a_{1}+2^{2}a_{2}+2^{3}a_{3}+\cdots +2^{n}a_{n}=n$,得$2a_{1}+2^{2}a_{2}+\cdots +2^{n-1}a_{n-1}=n-1(n\geq 2)$,两式相减,得$2^{n}a_{n}=1$,则$a_{n}=\frac{1}{2^{n}}(n\geq 2)$,又当$n=1$时,$2a_{1}=1$,解得$a_{1}=\frac{1}{2}$,满足上式,所以$a_{n}=\frac{1}{2^{n}}(n\in \mathbf{N}^{*})$,所以$b_{n}=\frac{1}{\log_{2}a_{n}\cdot \log_{2}a_{n+1}}=\frac{1}{\log_{2}2^{-n}\cdot \log_{2}2^{-(n+1)}}=\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}(n\in \mathbf{N}^{*})$,所以$S_{2021}=b_{1}+b_{2}+\cdots +b_{2021}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\cdots +\frac{1}{2021}-\frac{1}{2022}=1-\frac{1}{2022}=\frac{2021}{2022}$.故选C.

7. 记$S_{n}为等比数列\{ a_{n}\}$的前n项和. 若$a_{5}-a_{3}= 12,a_{6}-a_{4}= 24$,则$\frac {S_{n}}{a_{n}}$等于(B).
A.$2^{n}-1$
B.$2-2^{1-n}$
C.$2-2^{n-1}$
D.$2^{1-n}-1$

答案： B 解析:设等比数列$\{a_{n}\}$的公比为$q$,则$\frac{a_{6}-a_{4}}{a_{5}-a_{3}}=\frac{a_{5}\cdot q-a_{3}\cdot q}{a_{5}-a_{3}}=q=\frac{24}{12}=2$,所以$\frac{S_{n}}{a_{n}}=\frac{\frac{a_{1}(1-q^{n})}{1-q}}{a_{1}q^{n-1}}=\frac{1-q^{n}}{(1-q)q^{n-1}}=\frac{q-q^{n+1}}{(1-q)q^{n}}=\frac{2-2^{n+1}}{(1-2)2^{n}}=\frac{2^{n+1}-2}{2^{n}}=2-2^{1-n}$.故选B.

8. 在等比数列$\{ a_{n}\}$中,对任意$n∈N^{*},a_{1}+a_{2}+… +a_{n}= 2^{n}-1$,则$a_{1}^{2}+a_{2}^{2}+… +a_{n}^{2}$等于(D).
A.$(2^{n}-1)^{2}$
B.$\frac {(2^{n}-1)^{2}}{3}$
C.$4n-1$
D.$\frac {4^{n}-1}{3}$

答案： D 解析:因为$a_{1}+a_{2}+\cdots +a_{n}=2^{n}-1$,所以$a_{1}=2^{1}-1=1$.因为$a_{1}+a_{2}=1+a_{2}=2^{2}-1=3$,所以$a_{2}=2$,所以$\{a_{n}\}$的公比为2.所以$\{a_{n}^{2}\}$是等比数列,且公比为4,首项为$a_{1}^{2}=1$.所以$a_{1}^{2}+a_{2}^{2}+\cdots +a_{n}^{2}=\frac{a_{1}^{2}(1-4^{n})}{1-4}=\frac{4^{n}-1}{3}$.故选D.

9. 已知两个等差数列$\{ a_{n}\}和\{ b_{n}\}$的前n项和之比为$\frac {7n+1}{4n+27}(n∈N^{*})$,则$\frac {a_{11}}{b_{11}}= $$\frac{4}{3}$.

答案： $\frac{4}{3}$ 解析:设数列$\{a_{n}\}$的前$n$项和为$S_{n}$,数列$\{b_{n}\}$的前$n$项和为$T_{n}$,则$\frac{a_{11}}{b_{11}}=\frac{2a_{11}}{2b_{11}}=\frac{(a_{1}+a_{21})× 21}{(b_{1}+b_{21})× 21}=\frac{\frac{(a_{1}+a_{21})× 21}{2}}{\frac{(b_{1}+b_{21})× 21}{2}}=\frac{S_{21}}{T_{21}}=\frac{7× 21+1}{4× 21+27}=\frac{4}{3}$.

10. 已知函数$f(n)= \left\{\begin{array}{l} n^{2},n为奇数,\\ -n^{2},n为偶数,\end{array} \right.且a_{n}= f(n)+f(n+1)$,则$a_{1}+a_{2}+a_{3}+… +a_{100}= $100.

答案： 100 解析:由题意,得$a_{1}+a_{2}+a_{3}+\cdots +a_{100}=(1^{2}-2^{2})+(-2^{2}+3^{2})+(3^{2}-4^{2})+(-4^{2}+5^{2})+\cdots +(99^{2}-100^{2})+(-100^{2}+101^{2})=-(1+2)+(2+3)-\cdots -(99+100)+(100+101)=100$.

11. 已知数列$\{ a_{n}\}$是公差为d的等差数列,且$a_{1}= 2,a_{2}是a_{1},a_{4}$的等比中项.
(1)求数列$\{ a_{n}\}$的通项公式;
(2)当$d＞0$时,求数列$\{ \frac {1}{a_{n}(n+1)}\}$的前n项和$T_{n}$.

答案： 解:
(1)因为$a_{2}$是$a_{1},a_{4}$的等比中项,所以$(a_{1}+d)^{2}=a_{1}(a_{1}+3d)$,即$(2+d)^{2}=2(2+3d)$,整理,得$d^{2}-2d=0$,解得$d=0$或$d=2$.当$d=0$时,$a_{n}=2$;当$d=2$时,$a_{n}=2+2(n-1)=2n$.
(2)由
(1)知,当$d\gt 0$时,$a_{n}=2n$,所以$\frac{1}{a_{n}(n+1)}=\frac{1}{2n(n+1)}=\frac{1}{2}(\frac{1}{n}-\frac{1}{n+1})$,所以$T_{n}=\frac{1}{2}(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\cdots +\frac{1}{n}-\frac{1}{n+1})=\frac{1}{2}(1-\frac{1}{n+1})=\frac{n}{2n+2}$.