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 文（含解析）.docx

1、专题能力训练12数列的通项与求和一、能力突破训练1.已知数列an是等差数列,a1=tan 225,a5=13a1,设Sn为数列(-1)nan的前n项和,则S2 020=()A.2 020B.-2 020C.3 030D.-3 0302.已知数列an的前n项和为Sn,且Sn=n2+n,数列bn满足bn=1anan+1(nN*),Tn是数列bn的前n项和,则T9等于()A.919B.1819C.2021D.9403.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.1004.设数列an的前

2、n项和为Sn,且a1=1,Sn+nan为常数列,则an=()A.13n-1B.2n(n+1)C.1(n+1)(n+2)D.5-2n35.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()A.an=32-3213n,nN*B.an=32+3213n,nN*C.an=1,n=1,32+3213n,n2,且nN*D.an=1,nN*6.若数列an满足an+1=11-an,a11=2,则a1=.7.在数列an中,a2=5,an+1-an=2n(nN*),则数列an的通项公式an=.8.已知数列an满足3a1+32

3、a2+33a3+3nan=2n+1,则an的通项公式为.9.设数列an的前n项和为Sn.已知S2=4,an+1=2Sn+1,nN*.(1)求通项公式an;(2)求数列|an-n-2|的前n项和.10.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;(2)若数列1an2的前n项和为Tn,证明:Tnn+2,故bn=3n-1-n-2,n3.设数列bn的前n项和为Tn,则T1=2,T2=3.当n3时,Tn=3+9(1-3n-2)1-3-(n+7)(n-2)2=3n-n2-5n+112,所以Tn=2,n=1,3n-n2-5n+112,n2,nN*.10.(1)解当

4、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)=41-12+12-13+13-14+1n-1n+1=41-1n+14.

5、11.解(1)由a1=2,an+1=2an,得an=2n(nN*).由题意知,当n=1时,b1=b2-1,故b2=2.当n2时,1nbn=bn+1-bn,整理得bn+1n+1=bnn,所以bn=n(nN*).(2)由(1)知anbn=n2n,因此Tn=2+222+323+n2n,2Tn=22+223+324+n2n+1,所以Tn-2Tn=2+22+23+2n-n2n+1.故Tn=(n-1)2n+1+2(nN*).二、思维提升训练12.D解析:由题意,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+n

6、910n,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.-1n解析:由an+1=Sn+1-Sn=SnSn+1,得1Sn-1Sn+1=1,即1Sn+1-1Sn=-1,则1Sn为等差数列,首项为1S1=-1,公差为d=-1,1Sn=-n,Sn=-1n.14.(1)证明由条件,对任意nN*,有an+2=3Sn-Sn+1+3,因而对任意nN*,n2,有an+1=3Sn-1-Sn+3.两式相减,得an+2-

7、an+1=3an-an+1,即an+2=3an,n2.又a1=1,a2=2,所以a3=3S1-S2+3=3a1-(a1+a2)+3=3a1,故对一切nN*,an+2=3an.(2)解由(1)知,an0,所以an+2an=3,于是数列a2n-1是首项a1=1,公比为3的等比数列;数列a2n是首项a2=2,公比为3的等比数列.因此a2n-1=3n-1,a2n=23n-1.于是S2n=a1+a2+a2n=(a1+a3+a2n-1)+(a2+a4+a2n)=(1+3+3n-1)+2(1+3+3n-1)=3(1+3+3n-1)=3(3n-1)2,从而S2n-1=S2n-a2n=3(3n-1)2-23n-

8、1=32(53n-2-1).综上所述,Sn=32(53n-32-1),n是奇数,32(3n2-1),n是偶数.15.解(1)设数列an的公比为q.由已知,有1a1-1a1q=2a1q2,解得q=2或q=-1.又由S6=a11-q61-q=63,知q-1,所以a11-261-2=63,得a1=1.所以an=2n-1.(2)由题意,得bn=12(log2an+log2an+1)=12(log22n-1+log22n)=n-12,即bn是首项为12,公差为1的等差数列.设数列(-1)nbn2的前n项和为Tn,则T2n=(-b12+b22)+(-b32+b42)+(-b2n-12+b2n2)=b1+b

9、2+b3+b4+b2n-1+b2n=2n(b1+b2n)2=2n2.16.(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.