2021届高考数学二轮复习 专题能力训练12 数列的通项与求和 理（含解析）.docx

1、专题能力训练12数列的通项与求和专题能力训练第30页一、能力突破训练1.已知数列an是等差数列,满足a1+2a2=S5,下列结论错误的是()A.S9=0B.S5最小C.S3=S6D.a5=0答案:B解析:由题设可得3a1+2d=5a1+10d2a1+8d=0,即a5=0,所以D中结论正确.由等差数列的性质可得a1+a9=2a5=0,则S9=9(a1+a9)2=9a5=0,所以A中结论正确.S3-S6=3a1+3d-6a1-15d=-3(a1+4d)=-3a5=0,所以C中结论正确.B中结论是错误的.故选B.2.已知数列an的前n项和Sn=n2-2n-1,则a3+a17=()A.15B.17C.

2、34D.398答案:C解析:Sn=n2-2n-1,a1=S1=12-2-1=-2.当n2时,an=Sn-Sn-1=n2-2n-1-(n-1)2-2(n-1)-1=n2-(n-1)2+2(n-1)-2n-1+1=n2-n2+2n-1+2n-2-2n=2n-3.an=-2,n=1,2n-3,n2.a3+a17=(23-3)+(217-3)=3+31=34.3.在数列an中,an+1=2an-1,a3=2,设其前n项和为Sn,则S6=()A.874B.634C.15D.27答案:A解析:由an+1=2an-1,得an+1-1=2(an-1),则an-1是等比数列,首项为14,公比为2,则an-1=1

3、42n-1=2n-3,即an=1+2n-3,S6=6+14(1-26)1-2=874.4.已知数列an的前n项和为Sn,a1=1,a2=2,且对于任意n1,nN*,满足Sn+1+Sn-1=2(Sn+1),则S10的值为()A.90B.91C.96D.100答案:B解析:Sn+1+Sn-1=2(Sn+1),Sn+1-Sn=Sn-Sn-1+2,an+1-an=2.当n2时,数列an是等差数列,公差为2.又a1=1,a2=2,S10=1+92+9822=91.5.已知数列an,构造一个新数列a1,a2-a1,a3-a2,an-an-1,此数列是首项为1,公比为13的等比数列,则数列an的通项公式为(

4、)A.an=32-3213n,nN*B.an=32+3213n,nN*C.an=1,n=1,32+3213n,n2,且nN*D.an=1,nN*答案:A解析:因为数列a1,a2-a1,a3-a2,an-an-1,是首项为1,公比为13的等比数列,所以an-an-1=13n-1,n2.所以当n2时,an=a1+(a2-a1)+(a3-a2)+(an-an-1)=1+13+132+13n-1=1-13n1-13=32-3213n.又当n=1时,an=32-3213n=1,则an=32-3213n,nN*.6.已知数列an满足a1=1,an-an+1=nanan+1(nN*),则an=.答案:2n2

5、-n+2解析:因为an-an+1=nanan+1,所以an-an+1anan+1=1an+1-1an=n,1an=1an-1an-1+1an-1-1an-2+1a2-1a1+1a1=(n-1)+(n-2)+3+2+1+1a1=(n-1)(n-1+1)2+1=n2-n+22(n2).所以an=2n2-n+2(n2).又a1=1也满足上式,所以an=2n2-n+2.7.(2018全国,理14)记Sn为数列an的前n项和.若Sn=2an+1,则S6=.答案:-63解析:Sn=2an+1,Sn-1=2an-1+1(n2).-,得an=2an-2an-1,即an=2an-1(n2).又S1=2a1+1,

6、a1=-1.an是以-1为首项,2为公比的等比数列,则S6=-1(1-26)1-2=-63.8.已知Sn是等差数列an的前n项和,若a1=-2 017,S20142014-S20082008=6,则S2 017=.答案:-2 017解析:Sn是等差数列an的前n项和,Snn是等差数列,设其公差为d.S20142014-S20082008=6,6d=6,d=1.a1=-2017,S11=-2017.Snn=-2017+(n-1)1=-2018+n.S2017=(-2018+2017)2017=-2017.9.已知数列an的前n项和为Sn,且2Sn=nan+2an-1.(1)求数列an的通项公式;

7、(2)若数列1an2的前n项和为Tn,证明:Tn4.答案：(1)解当n=1时,2S1=a1+2a1-1,则a1=1.当n2时,2Sn=nan+2an-1,2Sn-1=(n-1)an-1+2an-1-1,-,得2an=nan-(n-1)an-1+2an-2an-1,即nan=(n+1)an-1,所以ann+1=an-1n,且a12=12,所以数列ann+1为常数列,ann+1=12,即an=n+12(nN*).(2)证明由(1)得an=n+12,所以1an2=4(n+1)24n(n+1)=41n-1n+1.所以Tn=422+432+442+4(n+1)2412+423+434+4n(n+1)=4

8、1-12+12-13+13-14+1n-1n+1=41-1n+11时,2Sn-1=3n-1+3,此时2an=2Sn-2Sn-1=3n-3n-1=23n-1,即an=3n-1,所以an=3,n=1,3n-1,n1.(2)因为anbn=log3an,所以b1=13,当n1时,bn=31-nlog33n-1=(n-1)31-n.所以T1=b1=13;当n1时,Tn=b1+b2+b3+bn=13+13-1+23-2+(n-1)31-n,所以3Tn=1+130+23-1+(n-1)32-n,两式相减,得2Tn=23+(30+3-1+3-2+32-n)-(n-1)31-n=23+1-31-n1-3-1-(

9、n-1)31-n=136-6n+323n,所以Tn=1312-6n+343n.经检验,当n=1时也适合.综上可得Tn=1312-6n+343n.二、思维提升训练12.“垛积术”(隙积术)是由北宋科学家沈括在梦溪笔谈中首创,南宋数学家杨辉、元代数学家朱世杰丰富和发展的一类数列求和方法,有菱草垛、方垛、刍童垛、三角垛等.如图,某仓库中部分货物堆放成“菱草垛”:自上而下,第一层1件,以后每一层比上一层多1件,最后一层是n件.已知第一层货物的单价是1万元,从第二层起,货物的单价是上一层单价的910.若这堆货物的总价是100-200910n万元,则n的值为()A.7B.8C.9D.10答案:D解析:由题

10、意,得第n层货物的总价为n910n-1万元.这堆货物的总价W=1+2910+39102+n910n-1,910W=1910+29102+39103+n910n,两式相减,得110W=-n910n+1+910+9102+9103+910n-1=-n910n+1-910n1-910=-n910n+10-10910n,则W=-10n910n+100-100910n=100-200910n,解得n=10.13.设Sn是数列an的前n项和,且a1=-1,an+1=SnSn+1,则Sn=.答案:-1n解析:由an+1=Sn+1-Sn=SnSn+1,得1Sn-1Sn+1=1,即1Sn+1-1Sn=-1,则1

11、Sn为等差数列,首项为1S1=-1,公差为d=-1,1Sn=-n,Sn=-1n.14.已知等差数列an的公差为2,其前n项和Sn=pn2+2n(nN*).(1)求p的值及an;(2)若bn=2(2n-1)an,记数列bn的前n项和为Tn,求使Tn910成立的最小正整数n的值.解：(1)(方法一)an是等差数列,Sn=na1+n(n-1)2d=na1+n(n-1)22=n2+(a1-1)n.又由已知Sn=pn2+2n,p=1,a1-1=2,a1=3,an=a1+(n-1)d=2n+1,p=1,an=2n+1.(方法二)由已知a1=S1=p+2,S2=4p+4,即a1+a2=4p+4,a2=3p+

12、2.又等差数列的公差为2,a2-a1=2,2p=2,p=1,a1=p+2=3,an=a1+(n-1)d=2n+1,p=1,an=2n+1.(方法三)当n2时,an=Sn-Sn-1=pn2+2n-p(n-1)2+2(n-1)=2pn-p+2,a2=3p+2,由已知a2-a1=2,2p=2,p=1,a1=p+2=3,an=a1+(n-1)d=2n+1,p=1,an=2n+1.(2)由(1)知bn=2(2n-1)(2n+1)=12n-1-12n+1,Tn=b1+b2+b3+bn=11-13+13-15+15-17+12n-1-12n+1=1-12n+1=2n2n+1.Tn910,2n2n+1910,

13、20n18n+9,即n92.nN*,使Tn910成立的最小正整数n的值为5.15.已知数列an满足an+2=qan(q为实数,且q1),nN*,a1=1,a2=2,且a2+a3,a3+a4,a4+a5成等差数列.(1)求q的值和an的通项公式;(2)设bn=log2a2na2n-1,nN*,求数列bn的前n项和.解：(1)由已知,有(a3+a4)-(a2+a3)=(a4+a5)-(a3+a4),即a4-a2=a5-a3,所以a2(q-1)=a3(q-1).又因为q1,故a3=a2=2,由a3=a1q,得q=2.当n=2k-1(kN*)时,an=a2k-1=2k-1=2n-12;当n=2k(kN

14、*)时,an=a2k=2k=2n2.所以,an的通项公式为an=2n-12,n为奇数,2n2,n为偶数.(2)由(1)得bn=log2a2na2n-1=n2n-1.设bn的前n项和为Sn,则Sn=1120+2121+3122+(n-1)12n-2+n12n-1,12Sn=1121+2122+3123+(n-1)12n-1+n12n,上述两式相减,得12Sn=1+12+122+12n-1-n2n=1-12n1-12-n2n=2-22n-n2n,整理得,Sn=4-n+22n-1.所以,数列bn的前n项和为4-n+22n-1,nN*.16.已知数列an,bn,Sn为数列an的前n项和,a1=2b1,

15、Sn=2an-2,nbn+1-(n+1)bn=n2+n.(1)求数列an的通项公式;(2)证明:数列bnn为等差数列;(3)若cn=-anbn2,n为奇数,anbn4,n为偶数,求数列cn的前2n项和.答案：(1)解Sn=2an-2,Sn-1=2an-1-2(n2),an=2an-2an-1(n2),an=2an-1(n2),数列an是以2为公比的等比数列.又a1=S1=2a1-2,a1=2,an=22n-1=2n.(2)证明nbn+1-(n+1)bn=n2+n=n(n+1),bn+1n+1-bnn=1,数列bnn是以1为公差的等差数列.(3)解b1=1,由(2)知bnn=b1+(n-1)1=n,bn=n2,cn=-n22n-1,n为奇数,n22n-2,n为偶数,c2n-1+c2n=-(2n-1)222n-2+(2n)222n-2=(4n-1)4n-1,T2n=340+741+(4n-1)4n-1,4T2n=341+742+(4n-5)4n-1+(4n-1)4n,-3T2n=3+441+42+4n-1-(4n-1)4n=3+16(1-4n-1)1-4-(4n-1)4n=4n+1-73-(4n-1)4n=-4n(12n-3)+4n+1-73,T2n=4n(12n-7)+79.