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1、课时规范练32数列求和基础巩固组1.设数列an为等比数列,数列bn为等差数列,且Sn为数列bn的前n项和,若a2=1,a10=16,且a6=b6,则S11=()A.20B.30C.44D.882.Sn=12+222+38+n2n等于()A.2n+1-n-22nB.2n-n2nC.2n-n+12n+1D.2n+1-n+22n3.(2022福建漳州二模)已知Sn是数列an的前n项和,a1=1,a2=2,a3=3,记bn=an+an+1+an+2且bn+1-bn=2,则S31=()A.171B.278C.351D.3954.在数列an中,an+an+1=2n,Sn为其前n项和,若a1=a4,则S10
2、1=()A.4 882B.5 100C.5 102D.5 2125.已知数列an的通项公式为an=n(n+1)!,则其前n项和为()A.1-1(n+1)!B.1-1n!C.2-1n!D.2-1(n+1)!6.(2022湖北荆州模拟)已知数列an满足an=1+2+4+2n-1,则数列2nanan+1的前5项和为.7.(2022辽宁辽阳二模)2nan为等差数列,且a3=58;an2n-1为等比数列,且a2=34.从两个条件中任选一个,补充在下面的问题中,并解答.在数列an中,a1=12,.(1)求an的通项公式;(2)已知an的前n项和为Sn,试问是否存在正整数p,q,r,使得Sn=p-qan+r
3、?若存在,求p,q,r的值;若不存在,请说明理由.8.已知数列an的前n项和Sn满足Sn=3n2-n2,数列log3bn是公差为-1的等差数列,b1=1.(1)求数列an,bn的通项公式;(2)设cn=a2n+1+b2n+1,求数列cn的前n项和Tn.综合提升组9.已知数列an的前n项和为Sn=2-3n,则此数列奇数项的前m项和为()A.949m4B.549m4C.949m-14D.-349m-1410.对于实数x,x表示不超过x的最大整数.已知数列an的通项公式为an=1n+1+n,前n项和为Sn,则S1+S2+S40=()A.105B.120C.125D.13011.在数列an中,a1=1
4、,nan+1=(n+1)an+n(n+1),若bn=ancos2n3,且数列bn的前n项和为Sn,则S11=()A.64B.80C.-64D.-8012.在数列an中,a1=2,ap+q=apaq(p,qN*),记bm为数列an中在区间(0,m(mN*)内的项的个数,则数列bm的前150项和S150=.13.(2022辽宁沈阳模拟)已知Sn为等差数列an的前n项和,a1+a5=22,Sn=n(an-2n+2).(1)求an的通项公式;(2)设bn=(-1)n8n+2anan+1,求数列bn的前2n+1项和T2n+1.14.在等差数列an中,a1+3,a3,a4成等差数列,且a1,a3,a8成等
5、比数列.(1)求数列an的通项公式;(2)在任意相邻两项ak与ak+1(k=1,2,)之间插入2k个2,使它们和原数列的项构成一个新的数列bn,求数列bn的前200项和T200.创新应用组15.在数列an中,a1=3,an+1=3an-4n,若bn=4n2+8n+5anan+1,且数列bn的前n项和为Sn,则Sn=()A.n1+22n+3B.43+2n6n+9C.n1+16n+9D.n1+26n+916.(2022山东潍坊二模)已知正项数列an的前n项和为Sn,且an2+2an=4Sn,数列bn满足bn=(-2)an2.(1)求数列bn的前n项和Bn,并证明Bn+1,Bn,Bn+2是等差数列;
6、(2)设cn=(-1)nan+bn,求数列cn的前n项和Tn.17.已知等比数列an的前n项和为Sn,公比q0,S2=2a2-2,S3=a4-2,数列an满足a2=4b1,nbn+1-(n+1)bn=n2+n(nN*).(1)求数列an的通项公式;(2)证明:数列bnn为等差数列;(3)设数列cn的通项公式为cn=-anbn2,n为奇数,anbn4,n为偶数,其前n项和为Tn,求T2n.课时规范练32数列求和1.C解析:数列an为等比数列,a62=a2a10=16.又a6=a2q40(q为公比),b6=a6=4.又数列bn为等差数列,S11=a1+a11211=11a6=44.故选C.2.A解
7、析:由Sn=12+222+323+n2n,可得12Sn=122+223+n-12n+n2n+1.两式相减可得,12Sn=12+122+123+12nn2n+1=121-(12)n1-12n2n+1=2n+1-n-22n+1,所以Sn=2n+1-n-22n,故选A.3.C解析:由bn+1-bn=2,得an+1+an+2+an+3-(an+an+1+an+2)=an+3-an=2,a1,a4,a7,是首项为1,公差为2的等差数列,a2,a5,a8,是首项为2,公差为2的等差数列,a3,a6,a9,是首项为3,公差为2的等差数列,S31=(a1+a4+a31)+(a2+a5+a29)+(a3+a6+
8、a30)=111+111022+210+10922+310+10922=351.4.C解析:因为an+an+1=2n,所以an+1+an+2=2n+2,由-得an+2-an=2,所以数列an奇数项与偶数项均成公差为2的等差数列.当n为奇数时,an=a1+n-122=n+a1-1;当n为偶数时,an=a2+n-222=n+a2-2=n+(2-a1)-2=n-a1.又因为a1=a4,所以a1=4-a1,得a1=2,所以an=n+1,n为奇数,n-2,n为偶数,所以S101=(a1+a101)+(a2+a100)=512(2+102)+502(0+98)=5102.故选C.5.A解析:因为an=n(
9、n+1)!=n+1-1(n+1)!=1n!1(n+1)!,所以其前n项和为1-12!+12!13!+1n!1(n+1)!=1-1(n+1)!.故选A.6.6263解析:因为an=1+2+4+2n-1=2n-1,an+1=2n+1-1,所以2nanan+1=2n(2n-1)(2n+1-1)=(2n+1-1)-(2n-1)(2n-1)(2n+1-1)=12n-112n+1-1.所以2nanan+1的前5项和为121-1122-1+122-1123-1+125-1126-1=121-1126-1=1-163=6263.7.解 (1)若选:设等差数列2nan的公差为d,则d=23a3-2a13-1=5
10、-12=2,2nan=2a1+2(n-1)=2n-1,即an=2n-12n.若选:设等比数列an2n-1的公比为q,则q=a222-1a121-1=12,an2n-1=a121-112n-1=12n,即an=2n-12n.(2)Sn=12+322+2n-12n,12Sn=122+323+2n-12n+1,则两式相减得12Sn=12+2122+123+12n-2n-12n+1=12+2(14-12n+1)1-122n-12n+1=322n+32n+1,Sn=3-2n+32n.Sn=3-2n+32n=3-42(n+2)-12n+2=3-4an+2,存在正整数p,q,r,使得Sn=p-qan+r,且
11、p=3,q=4,r=2.8.解(1)当n=1时,a1=S1=3-12=1,当n2,an=Sn-Sn-1=3n2-n23(n-1)2-(n-1)2=3n-2,所以n=1满足n2时的情况,所以an=3n-2(nN*).因为log3bn=log3b1+(n-1)(-1)=1-n,所以bn=31-n.(2)因为cn=a2n+1+b2n+1=3(2n+1)-2+31-(2n+1)=6n+1+19n,所以Tn=(7+6n+1)n2+191-(19)n1-19,所以Tn=3n2+4n+181-19n.9.B解析:当n2时,an=Sn-Sn-1=(2-3n)-(2-3n-1)=-23n-1.因为当n=1时,a
12、1=-1不满足,所以数列an从第2项开始成等比数列.又a3=-18,则数列an的奇数项构成的数列的前m项和Tm=-18(1-9m-1)1-9-1=549m4.故选B.10.B解析:因为an=1n+1+n=n+1n,所以Sn=21+32+n+1n=n+1-1.S1=1+1-1=0,S2=2+1-1=0,S3=3+1-1=1,S4=4+1-1=1,S7=7+1-1=1,S8=8+1-1=2,S9=9+1-1=2,S14=14+1-1=2,S15=15+1-1=3,S16=16+1-1=3,S23=23+1-1=3,S24=24+1-1=4,S25=25+1-1=4,S34=34+1-1=4,S35
13、=35+1-1=5,S36=36+1-1=5,S40=40+1-1=5.故S1+S2+S40=02+15+27+39+411+56=120.故选B.11.C解析:已知nan+1=(n+1)an+n(n+1),则an+1n+1=ann+1,可得数列ann是首项为1,公差为1的等差数列,即有ann=n,即an=n2,则bn=ancos2n3=n2cos2n3,则S11=-12(12+22+42+52+72+82+102+112)+(32+62+92)=-12(12+22-32-32+42+52-62-62+72+82-92-92+102+112)=-12(5+23+41+59)=-64.故选C.1
14、2.803解析:令p=1,q=n,则a1+n=a1an=2an,所以数列an是首项为2,公比为2的等比数列,所以an=2n.当m=1时,b1=0.当2nm0,当n=1时,a12+2a1=4a1,a1=2或a1=0(舍去),当n2时,an-12+2an-1=4Sn-1,-,得an2an-12+2an-2an-1=4an,(an-an-1)(an+an-1)=2(an+an-1).an0,an-an-1=2(n2),an是以2为首项,2为公差的等差数列,an=2n,bn=(-2)n,数列bn是首项为-2,公比为-2的等比数列,Bn=-21-(-2)n1-(-2)=-23+(-1)n2n+13.Bn
15、+2+Bn+1=-43+(-1)n+22n+33+(-1)n+12n+23=-43+(-1)n+12n+2(-2+1)3=-43+2(-1)n2n+13=2Bn,Bn+1,Bn,Bn+2成等差数列.(2)cn=(-1)n2n+(-2)n=(-2)n+2(-1)nn,当n为偶数时,Tn=(-2)1+(-2)2+(-2)n+2-1+2-3+4+-(n-1)+n=-21-(-2)n1-(-2)+2(-1+2)+(-3+4)+(-n+1+n)=-2+2n+13+2n2=2n+13+n-23.当n为奇数时,Tn=(-2)1+(-2)2+(-2)n+2-1+2-3+4+(n-1)-n=-21-(-2)n1
16、-(-2)+2-1+(2-3)+(4-5)+(n-1-n)=-2-(-2)n+13+2-1+(-1)n-12=-2-2n+13-n-1=-2n+13-n-53.综上可知Tn=2n+13+n-23,n为偶数,-2n+13-n-53,n为奇数.17.(1)解因为等比数列an的前n项和为Sn,公比q0,S2=2a2-2,S3=a4-2,所以S3-S2=a4-2a2=a3,整理得a2q2-2a2=a2q.又a20,所以q2-q-2=0.由于q0,解得q=2.由于a1+a2=2a2-2,解得a1=2,所以an=2n.(2)证明数列an满足a2=4b1,解得b1=1.由于nbn+1-(n+1)bn=n2+
17、n,所以bn+1n+1bnn=1,所以数列bnn是以1为首项,1为公差的等差数列.(3)解因为数列bnn是以1为首项,1为公差的等差数列,所以bnn=1+(n-1)=n,所以bn=n2.因为数列cn的通项公式为cn=-anbn2,n为奇数,anbn4,n为偶数,所以令pn=c2n-1+c2n=-(2n-1)222n-12+(2n)222n4=(4n-1)4n-1,所以T2n=340+741+1142+(4n-1)4n-1,4T2n=341+742+1143+(4n-5)4n-1+(4n-1)4n.-得-3T2n=340+441+44n-1-(4n-1)4n,整理得-3T2n=3+44n-44-1-(4n-1)4n,故T2n=79+12n-794n.