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1、专题对点练13等差、等比数列与数列的通项及求和1.Sn为数列an的前n项和.已知an>0,an2+2an=4Sn+3.(1)求an的通项公式;(2)设bn=1anan+1,数列bn的前n项和为Tn,求Tn.解 (1)由an2+2an=4Sn+3,可知an+12+2an+1=4Sn+1+3.两式相减可得an+12-an2+2(an+1-an)=4an+1,即2(an+1+an)=an+12-an2=(an+1+an)(an+1-an).由于an>0,因此an+1-an=2.又a12+2a1=4a1+3,解得a1=3(a1=-1舍去).所以an是首项为3,公差为2的等差数列,故an=2
2、n+1.(2)由an=2n+1可知bn=1anan+1=1212n+1-12n+3.Tn=b1+b2+bn=1213-15+15-17+12n+1-12n+3=n3(2n+3).2.已知数列an是等差数列,前n项和为Sn,若a1=9,S3=21.(1)求数列an的通项公式;(2)若a5,a8,Sk成等比数列,求k的值.解 (1)设等差数列an的公差为d,a1=9,S3=21,S3=3×9+3×22d=21,解得d=-2,an=9+(n-1)×(-2)=-2n+11.(2)a5,a8,Sk成等比数列,a82=a5·Sk,即(-2×8+11)2=(
3、-2×5+11)·9k+k(k-1)2×(-2),解得k=5.3.(2017河北衡水中学三调,理17)已知数列an的前n项和为Sn,a10,常数>0,且a1an=S1+Sn对一切正整数n都成立.(1)求数列an的通项公式;(2)设a1>0,=100,当n为何值时,数列lg1an的前n项和最大?解 (1)令n=1,得a12=2S1=2a1,即a1(a1-2)=0.因为a10,所以a1=2.当n2时,2an=2+Sn,2an-1=2+Sn-1,两式相减,得2an-2an-1=an(n2),所以an=2an-1(n2),从而数列an为等比数列,所以an=a1
4、·2n-1=2n.(2)当a1>0,=100时,由(1)知,an=2n100,设bn=lg1an=lg1002n=lg 100-lg 2n=2-nlg 2,所以数列bn是单调递减的等差数列,公差为-lg 2,所以b1>b2>>b6=lg10026=lg10064>lg 1=0,当n7时,bnb7=lg10027<lg 1=0,所以数列lg1an的前6项和最大.4.(2017河北邯郸二模,理17)已知等差数列an的前n项和为Sn,a10,a3=3,且Sn=anan+1.在等比数列bn中,b1=2,b3=a15+1.(1)求数列an及bn的通项公式;(
5、2)设数列cn的前n项和为Tn,且Sn+n2cn=1,求Tn.解 (1)Sn=anan+1,a3=3,a1=a1a2,且(a1+a2)=a2a3,a2=,a1+a2=a3=3.数列an是等差数列,a1+a3=2a2,即2a2-a1=3.由得a1=1,a2=2,an=n,=2,b1=4,b3=16,bn的公比q=±b3b1=±2,bn=2n+1或bn=(-2)n+1.(2)由(1)知Sn=n(1+n)2,cn=2n(n+2)=1n-1n+2,Tn=1-13+12-14+13-15+1n-1-1n+1+1n-1n+2=1+12-1n+1-1n+2=32-2n+3n2+3n+2.
6、5.(2017宁夏中卫二模,理17)已知等比数列an的公比q>1,且a1+a3=20,a2=8.(1)求数列an的通项公式;(2)设bn=nan,Sn是数列bn的前n项和,求Sn.解 (1)等比数列an的公比q>1,且a1+a3=20,a2=8,a1+a1q2=20,a1q=8,2q2-5q+2=0,解得q=2,a1=4.an=2n+1.(2)bn=nan=n2n+1,Sn=122+223+n2n+1,12Sn=123+224+n-12n+1+n2n+2.12Sn=122+123+12n+1-n2n+2=141-12n1-12-n2n+2=12-2+n2n+2.Sn=1-2+n2n
7、+1.6.(2017安徽安庆二模,理17)在数列an中,a1=2,a2=4,设Sn为数列an的前n项和,对于任意的n>1,nN*,Sn+1+Sn-1=2(Sn+1).(1)求数列an的通项公式;(2)设bn=n2an,求bn的前n项和Tn.解 (1)对于任意的n>1,nN*,Sn+1+Sn-1=2(Sn+1),Sn+2+Sn=2(Sn+1+1),两式相减可得an+2+an=2an+1.(*)当n=2时,S3+S1=2(S2+1),即2a1+a2+a3=2(a1+a2+1),解得a3=6.当n=1时(*)也满足.数列an是等差数列,公差为2,an=2+2(n-1)=2n.(2)bn=
8、n2an=n22n=n4n,Tn=14+242+343+n4n,14Tn=142+243+n-14n+n4n+1,34Tn=14+142+14n-n4n+1=141-14n1-14-n4n+1,Tn=49-4+3n9×4n.导学号168041897.(2017山东,理19)已知xn是各项均为正数的等比数列,且x1+x2=3,x3-x2=2.(1)求数列xn的通项公式;(2)如图,在平面直角坐标系xOy中,依次连接点P1(x1,1),P2(x2,2)Pn+1(xn+1,n+1)得到折线P1P2Pn+1,求由该折线与直线y=0,x=x1,x=xn+1所围成的区域的面积Tn.解 (1)设数
9、列xn的公比为q,由已知q>0.由题意得x1+x1q=3,x1q2-x1q=2.所以3q2-5q-2=0.因为q>0,所以q=2,x1=1,因此数列xn的通项公式为xn=2n-1.(2)过P1,P2,Pn+1向x轴作垂线,垂足分别为Q1,Q2,Qn+1.由(1)得xn+1-xn=2n-2n-1=2n-1,记梯形PnPn+1Qn+1Qn的面积为bn,由题意bn=(n+n+1)2×2n-1=(2n+1)×2n-2,所以Tn=b1+b2+bn=3×2-1+5×20+7×21+(2n-1)×2n-3+(2n+1)×2n-
10、2.又2Tn=3×20+5×21+7×22+(2n-1)×2n-2+(2n+1)×2n-1,-得-Tn=3×2-1+(2+22+2n-1)-(2n+1)×2n-1=32+2(1-2n-1)1-2-(2n+1)×2n-1.所以Tn=(2n-1)×2n+12.导学号168041908.(2017山东潍坊一模,理19)已知数列an是等差数列,其前n项和为Sn,数列bn是公比大于0的等比数列,且b1=-2a1=2,a3+b2=-1,S3+2b3=7.(1)求数列an和bn的通项公式;(2)令cn=2,n为奇数,-
11、2anbn,n为偶数,求数列cn的前n项和Tn.解 (1)设等差数列an的公差为d,等比数列bn的公比为q,q>0,b1=-2a1=2,a3+b2=-1,S3+2b3=7,a1=-1,b1=2,-1+2d+2q=-1,3×(-1)+3d+2×2×q2=7,解得d=-2,q=2.an=-1-2(n-1)=1-2n,bn=2n.(2)cn=2,n为奇数,2n-12n-1,n为偶数.当n=2k(kN*)时,数列cn的前n项和Tn=T2k=(c1+c3+c2k-1)+(c2+c4+c2k)=2k+32+723+4k-122k-1,令Ak=32+723+4k-122k
12、-1,14Ak=323+725+4k-522k-1+4k-122k+1,34Ak=32+4123+125+122k-1-4k-122k+1=32+4×181-14k-11-14-4k-122k+1,Ak=269-12k+139×22k-1.Tn=T2k=2k+269-12k+139×22k-1.当n=2k-1(kN*)时,数列cn的前n项和Tn=T2k-2+a2k-1=2(k-1)+269-12(k-1)+139×22(k-1)-1+2=2k+269-12k+19×22k-3.Tn=2k+269-12k+139×22k-1,n=2k,2k+269-12k+19×22k-3,n=2k-1,kN*.6EDBC3191F2351DD815FF33D4435F3756EDBC3191F2351DD815FF33D4435F3756EDBC3191F2351DD815FF33D4435F3756EDBC3191F2351DD815FF33D4435F3756EDBC3191F2351DD815FF33D4435F3756EDBC3191F2351DD815FF33D4435F375
高考数学二轮复习 专题对点练13 等差、等比数列与数列的通项及求和 理
