2021-2022学年数学北师大版选择性必修第二册测评：第一章　习题课2　数列的求和问题 WORD版含解析.docx

1、第一章数列习题课2数列的求和问题课后篇巩固提升必备知识基础练1.已知数列an的通项an=2n+1,nN+,由bn=a1+a2+a3+ann所确定的数列bn的前n项的和是()A.n(n+2)B.12n(n+4)C.12n(n+5)D.12n(n+7)答案C解析a1+a2+an=n2(2n+4)=n2+2n,bn=n+2,bn的前n项和Sn=n(n+5)2.2.数列125,158,1811,1(3n-1)(3n+2),的前n项和为()A.n3n+2B.n6n+4C.3n6n+4D.n+1n+2答案B解析由数列通项公式1(3n-1)(3n+2)=1313n-1-13n+2,得前n项和Sn=1312-

2、15+15-18+18-111+13n-1-13n+2=1312-13n+2=n6n+4.3.1+1+12+1+12+14+1+12+14+1210的值为()A.18+129B.20+1210C.22+1211D.18+1210答案B解析设an=1+12+14+12n-1=11-(12)n1-12=21-12n,原式=a1+a2+a11=21-121+21-122+21-1211=211-12+122+1211=211-12(1-1211)1-12=210+1211=20+1210.4.设an=1n+1+n,数列an的前n项和Sn=9,则n=.答案99解析an=1n+1+n=n+1-n,故Sn

3、=2-1+3-2+n+1-n=n+1-1=9.解得n=99.5.在数列an中,已知Sn=1-5+9-13+17-21+(-1)n-1(4n-3),nN+,则S15+S22-S31的值是.答案-76解析S15=-47+a15=-28+57=29,S22=-411=-44,S31=-415+a31=-60+121=61,S15+S22-S31=29-44-61=-76.6.已知函数f(x)=2x-3x-1,点(n,an)(nN+)在f(x)的图象上,数列an的前n项和为Sn,求Sn.解由题意得an=2n-3n-1,nN+,Sn=a1+a2+an=(2+22+2n)-3(1+2+3+n)-n=2(1

4、-2n)1-2-3n(n+1)2-n=2n+1-n(3n+5)2-2.7.已知等差数列an中,2a2+a3+a5=20,且前10项和S10=100.(1)求数列an的通项公式;(2)若bn=1anan+1,求数列bn的前n项和.解(1)设数列an的通项公式为an=a1+(n-1)d,由已知得2a2+a3+a5=4a1+8d=20,10a1+1092d=10a1+45d=100,解得a1=1,d=2,所以数列an的通项公式为an=1+2(n-1)=2n-1.(2)bn=1(2n-1)(2n+1)=1212n-1-12n+1,所以Tn=121-13+13-15+12n-1-12n+1=121-12

5、n+1=n2n+1.关键能力提升练8.已知函数f(x)=21+x2(xR),若等比数列an满足a1a2 021=1,则f(a1)+f(a2)+f(a3)+f(a2 021)=()A.2 021B.20212C.2D.12答案A解析函数f(x)=21+x2(xR),f(x)+f1x=21+x2+21+(1x)2=21+x2+2x2x2+1=2.数列an为等比数列,且a1a2021=1,a1a2021=a2a2020=a3a2019=a2021a1=1.f(a1)+f(a2021)=f(a2)+f(a2020)=f(a3)+f(a2019)=f(a2021)+f(a1)=2,f(a1)+f(a2)

6、+f(a3)+f(a2021)=2021.故选A.9.已知等差数列an中,a3+a5=a4+7,a10=19,则数列ancos n的前2 022项和为()A.1 010B.1 011C.2 021D.2 022答案D解析设数列an的公差为d,由2a1+6d=a1+3d+7,a1+9d=19,解得a1=1,d=2,an=2n-1.设bn=ancosn,b1+b2=a1cos+a2cos2=2,b3+b4=a3cos3+a4cos4=2,数列ancosn的前2022项的和S2022=(b1+b2)+(b3+b4)+(b2021+b2022)=10112=2022.故选D.10.定义np1+p2+p

7、n为n个正数p1,p2,pn的“均倒数”.若已知数列an的前n项的“均倒数”为13n+1,bn=an+26,则1b1b2+1b2b3+1b9b10=()A.111B.1011C.910D.1112答案C解析由题意得na1+a2+an=13n+1,所以a1+a2+an=n(3n+1)=3n2+n,记数列an的前n项和为Sn,则Sn=3n2+n.当n=1时,a1=S1=4;当n2时,an=Sn-Sn-1=3n2+n-3(n-1)2+(n-1)=6n-2.经检验a1=4也符合此式,所以an=6n-2,nN+,则bn=an+26=n,所以1b1b2+1b2b3+1b9b10=112+123+1910=

8、1-12+12-13+19-110=1-110=910.故选C.11.(多选题)记Sn为等差数列an的前n项和,若S5=0,a6=6,则()A.an=2n-6B.an=3n-12C.Sn=n2-5nD.Sn=n2-5n2答案AC解析设数列an的公差是d,Sn为等差数列an的前n项和,S5=0,a6=6,S5=5a1+542d=0,a6=a1+5d=6,解得a1=-4,d=2,an=-4+(n-1)2=2n-6,Sn=-4n+n(n-1)22=n2-5n.故A和C正确,B和D错误.12.(多选题)已知数列an满足a1=1,an+1=lg(10an+9)+1,其前n项和为Sn,则下列结论中正确的有

9、()A.an是递增数列B.an+10是等比数列C.2an+1an+an+2D.Sn2102n+2=210n+1,所以2an+1an+an+2,C正确;令cn=n+1,则其前n项和为n(n+3)2,而an=lg(210n-10)lg(210n)lg(1010n)=n+1=cn,故Snn(n+3)2,D正确.13.已知数列an的前n项和为Sn,且满足Sn=2an-1(nN+),则数列nan的前n项和Tn为.答案(n-1)2n+1解析Sn=2an-1(nN+),n=1时,a1=2a1-1,解得a1=1,n2时,an=Sn-Sn-1=2an-1-(2an-1-1),化为an=2an-1,数列an是首项

10、为1,公比为2的等比数列,an=2n-1.nan=n2n-1.则数列nan的前n项和Tn=1+22+322+n2n-1.2Tn=2+222+(n-1)2n-1+n2n,-Tn=1+2+22+2n-1-n2n=1-2n1-2-n2n=(1-n)2n-1,Tn=(n-1)2n+1.14.设等差数列an满足a2=5,a6+a8=30,则an=,数列1an2-1的前n项和为.答案2n+1n4(n+1)解析设数列an的公差为d.a6+a8=30=2a7,a7=15,a7-a2=5d,a2=5,d=2,an=a2+(n-2)d=2n+1.1an2-1=14n(n+1)=141n-1n+1,1an2-1的前

11、n项和为141-12+12-13+1n-1n+1=141-1n+1=n4(n+1).15.等差数列an中,a2=4,a5+a6=15.(1)求数列an的通项公式;(2)设bn=2an-2+n-8,求b1+b2+b3+b10的值.解(1)设数列an的公差为d,a5+a6=15,a2+a9=15,又a2=4,a9=11,d=a9-a29-2=11-47=1,an=a2+(n-2)1=n+2.(2)由(1)可得bn=2an-2+n-8=2n+n-8,b1+b2+b3+b10=(2+22+23+210)+(1+2+3+10)-80=2(1-210)1-2+10(1+10)2-80=211-2+55-8

12、0=2021.学科素养创新练16.已知正项数列an的首项a1=1,其前n项和为Sn,且an与an+1的等比中项是2Sn,数列bn满足:b1+b2+bn=an2an+2.(1)求a2,a3,并求数列an的通项公式;(2)记cn=bnan,nN+,证明:c1+c2+cn21-1n+1.(1)解由an与an+1的等比中项是2Sn,得2Sn=anan+1,分别取n=1,2,得2a1=a1a2,2(a1+a2)=a2a3,解得a2=2,a3=3.又2Sn+1=an+1an+2,联立可得an+2-an=2.又a1=1,a2=2,an=n.(2)证明依题意,b1+b2+bn=an2an+2=n2(n+2)=12-1n+2,当n2时,b1+b2+bn-1=12-1n+1,两式相减即得bn=1n+1-1n+2=1(n+1)(n+2).b1=a12a3=16,符合上式,bn=1(n+1)(n+2),则cn=bnan=1n(n+1)(n+2)=2n(n+1)(n+2+n+2)2n(n+1)(n+n+1)=2(n+1-n)n(n+1)=21n-1n+1.c1+c2+cn21-12+12-13+1n-1n+1=21-1n+1.