2021届高中数学统考第二轮专题复习 第10讲 数列求和与数列的简单应用限时集训（理含解析）.docx

1、第10讲 数列求和与数列的简单应用基础过关1.在等比数列an中,a1=6,a2=12-a3.(1)求an的通项公式;(2)记Sn为an的前n项和,若Sm=66,求m.2.已知数列an是公比为正数的等比数列,其前n项和为Sn,满足a1=2,且a2,2S2,a3成等差数列.(1)求an的通项公式;(2)若数列bn满足bn=log2an,求b12-b22+b32-b42+b992-b1002的值.3.已知数列an,bn满足a1=b1=1,对任意nN*均有an+1=an+bn+an2+bn2,bn+1=an+bn-an2+bn2.(1)证明:数列an+bn和数列anbn均为等比数列;(2)设cn=(n

2、+1)2n1an+1bn,求数列cn的前n项和Tn.4.如图X10-1,杨辉三角是我国南宋数学家杨辉于1261年所著的详解九章算法中列出的一张图表,如图,把杨辉三角左对齐排列,将同一条斜线上的数字求和,会得到一个数列an,其中a1=1,a2=1,a3=2,设数列an的前n项和为Sn.图X10-1(1)求a8的值,并写出an,an+1,an+2满足的递推关系式(不用证明);(2)记a2022=m,用m表示S2020.5.已知数列an的前n项和为Sn,且2Sn=3an+4n-7.(1)证明:数列an-2为等比数列;(2)若bn=an-2(an+1-1)(an-1),求数列bn的前n项和Tn.6.已

3、知数列an的前n项和为Sn,且满足Sn=2an-1(nN*).(1)求数列an的通项公式及Sn;(2)若数列bn满足bn=|Sn-15|,求数列bn的前n项和Tn.能力提升7.已知数列an的首项a1=2,且满足a1+a2+an-an+1=-2,数列bn满足bn=1nlog2(a1a2an)(nN*),数列anbn的前n项和为Tn.(1)求数列an的通项公式;(2)令cn=n4Tn-2n,求证:i=1nci13.8.已知等比数列an的各项均为正数,2a5,a4,4a6成等差数列,且满足a4=4a32,数列bn的前n项和Sn=(n+1)2bn,nN*,且b1=1.(1)求数列an和bn的通项公式;

4、(2)设cn=bn,n为奇数,an,n为偶数,求数列cn的前n项和Pn.(3)设dn=b2n+5b2n+1b2n+3an,nN*,dn的前n项和为Tn,求证:Tn0).a2,2S2,a3成等差数列,4S2=a2+a3,即4(a1+a1q)=a1q+a1q2,4+4q=q+q2,解得q=-1或q=4,q0,q=4,a1=2,an=24n-1=22n-1(nN*).(2)由(1)知bn=log2an=2n-1,b12-b22+b32-b42+b992-b1002=12-32+52-72+1972-1992=(1-3)(1+3)+(5-7)(5+7)+(197-199)(197+199)=(-2)(

5、1+3+5+7+199)=-21+1992100=-20000.3.解:(1)证明:由an+1=an+bn+an2+bn2,bn+1=an+bn-an2+bn2,得an+1+bn+1=2(an+bn),又因为a1+b1=2,所以数列an+bn是首项为2,公比为2的等比数列.由an+1=an+bn+an2+bn2,bn+1=an+bn-an2+bn2,得an+1bn+1=2anbn,又因为a1b1=1,所以数列anbn是首项为1,公比为2的等比数列.(2)由(1)可知an+bn=2n,anbn=2n-1,所以1an+1bn=an+bnanbn=2,故cn=(n+1)2n1an+1bn=(n+1)

6、2n+1,所以Tn=222+323+424+(n+1)2n+1,则2Tn=223+324+425+(n+1)2n+2,两式相减得-Tn=222+23+24+2n+1-(n+1)2n+2,即-Tn=8+23(1-2n-1)1-2-(n+1)2n+2=-n2n+2,所以Tn=n2n+2.4.解:(1)由题图知a8=1+6+10+4=21,an+2=an+1+an(nN*).(2)因为a3=a2+a1,a4=a3+a2,a2021=a2020+a2019,a2022=a2021+a2020,所以a3+a4+a2022=a2+a3+a2021+(a1+a2+a2020),所以a2022-a2=S202

7、0,所以S2020=m-1.5.解:(1)证明:当n=1时,2a1=3a1-3,则a1=3.当n2时,因为2Sn=3an+4n-7,所以2Sn-1=3an-1+4n-11,则2an=3(an-an-1)+4,即an=3an-1-4,从而an-2=3(an-1-2),即an-2an-1-2=3(n2).因为a1=3,所以a1-2=1,所以数列an-2是以1为首项,3为公比的等比数列.(2)由(1)可得an-2=3n-1,即an=3n-1+2,所以bn=an-2(an+1-1)(an-1)=3n-1(3n+1)(3n-1+1)=1213n-1+1-13n+1,则Tn=12130+1-131+1+1

8、31+1-132+1+13n-2+1-13n-1+1+13n-1+1-13n+1=12130+1-13n+1=1212-13n+1=14-12(3n+1).6.解:(1)当n=1时,S1=2a1-1,即a1=1,由Sn=2an-1得Sn+1=2an+1-1,两式相减得an+1=2an+1-2an,即an+1=2an,数列an是以1为首项,2为公比的等比数列,an=2n-1,则Sn=1-2n1-2=2n-1.(2)由(1)知bn=|2n-16|,则bn=16-2n,1n4,2n-16,n4,当1n4时,Tn=(16-21)+(16-22)+(16-2n)=16n-(21+22+2n)=16n-2

9、(1-2n)1-2=16n-2n+1+2;当n4时,Tn=(16-21)+(16-22)+(16-23)+(16-24)+(25-16)+(26-16)+(27-16)+(2n-16)=2T4+(21+22+2n)-16n=234+2(1-2n)1-2-16n=2n+1-16n+66.综上,Tn=16n-2n+1+2,1n4,2n+1-16n+66,n4.7.解:(1)由a1=2,a1+a2+an-an+1=-2,可得a1-a2=-2,即a2=4.当n2时,a1+a2+an-1-an=-2,又a1+a2+an-an+1=-2,两式相减可得an+1=2an,又a2=4=2a1,满足上式,所以an

10、是首项为2,公比为2的等比数列,则an=22n-1=2n.(2)证明:由(1)知bn=1nlog2(2482n)=1+2+nn=n+12,则anbn=(n+1)2n-1,所以Tn=21+32+44+(n+1)2n-1,2Tn=22+34+48+(n+1)2n,两式相减可得-Tn=2+2+4+2n-1-(n+1)2n=2+2(1-2n-1)1-2-(n+1)2n,所以Tn=n2n,可得cn=n4Tn-2n=n4n2n-2n=142n-2.由42n-2-32n=2n-20,得42n-232n,可得142n-21312n,则i=1nci1312+14+12n=1312(1-12n)1-12=131-

11、12n0,所以q0,由题知2a4=2a5+4a6,a4=4a32,得2q2+q-1=0,1=4a1q,解得q=12,a1=12,所以an=12n.当n2时,bn=Sn-Sn-1=n+12bn-nbn-12,即bnn=bn-1n-1,所以bnn是常数列,又b11=1,所以bn=n.(2)由(1)知cn=n,n为奇数,(12)n,n为偶数.当n为偶数时,Pn=(b1+b3+bn-1)+(a2+a4+an)=1+3+(n-1)+122+124+12n=n22(1+n-1)+141-(14)n21-14=n24+13-1312n;当n为奇数时,Pn=Pn-1+bn=(n-1)24+13-1312n-1+n=(n+1)24+13-1312n-1.(3)证明:由(1)知dn=2n+5(2n+1)(2n+3)12n=1(2n+1)2n-1-1(2n+3)2n,则Tn=13-152+152-1722+1(2n+1)2n-1-1(2n+3)2n=13-1(2n+3)2n13.