新课标广西2022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

专题对点练13　等差、等比数列与数列的通项及求和1.已知各项都为正数的数列{an}满足a1=1,an2-(2an+1-1)·an-2an+1=0.(1)求a2,a3;(2)求{an}的通项公式.2.(2022北京,文15)设{an}是等差数列,且a1=ln2,a2+a3=5ln2.(1)求{an}的通项公式;(2)求ea1+ea2+…+ean.3.(2022全国Ⅲ,文17)等比数列{an}中,a1=1,a5=4a3.(1)求{an}的通项公式;(2)记Sn为{an}的前n项和,若Sm=63,求m.4.在等差数列{an}中,a2+a7=-23,a3+a8=-29.(1)求数列{an}的通项公式;(2)设数列{an+bn}是首项为1,公比为2的等比数列,求{bn}的前n项和Sn.4\n5.(2022天津,文18)设{an}是等差数列,其前n项和为Sn(n∈N*);{bn}是等比数列,公比大于0,其前n项和为Tn(n∈N*).已知b1=1,b3=b2+2,b4=a3+a5,b5=a4+2a6.(1)求Sn和Tn;(2)若Sn+(T1+T2+…+Tn)=an+4bn,求正整数n的值.6.在等差数列{an}中,a7=8,a19=2a9.(1)求{an}的通项公式;(2)设bn=1nan,求数列{bn}的前n项和Sn.7.已知{an}是各项均为正数的等比数列,且a1+a2=6,a1a2=a3.(1)求数列{an}的通项公式;(2){bn}为各项非零的等差数列,其前n项和为Sn.已知S2n+1=bnbn+1,求数列bnan的前n项和Tn.8.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)设cn=(-1)n-1anbn,求数列{cn}的前n项和Tn.4\n专题对点练13答案1.解(1)由题意得a2=12,a3=14.(2)由an2-(2an+1-1)an-2an+1=0得2an+1(an+1)=an(an+1).因为{an}的各项都为正数,所以an+1an=12.故{an}是首项为1,公比为12的等比数列,因此an=12n-1.2.解(1)设等差数列{an}的公差为d,∵a2+a3=5ln2,∴2a1+3d=5ln2.又a1=ln2,∴d=ln2.∴an=a1+(n-1)d=nln2.(2)由(1)知an=nln2.∵ean=enln2=eln2n=2n,∴{ean}是以2为首项,2为公比的等比数列.∴ea1+ea2+…+ean=2+22+…+2n=2n+1-2.∴ea1+ea2+…+ean=2n+1-2.3.解(1)设{an}的公比为q,由题设得an=qn-1.由已知得q4=4q2,解得q=0(舍去),q=-2或q=2.故an=(-2)n-1或an=2n-1.(2)若an=(-2)n-1,则Sn=1-(-2)n3.由Sm=63得(-2)m=-188,此方程没有正整数解.若an=2n-1,则Sn=2n-1.由Sm=63得2m=64,解得m=6.综上,m=6.4.解(1)设等差数列{an}的公差是d.由已知(a3+a8)-(a2+a7)=2d=-6,解得d=-3,∴a2+a7=2a1+7d=-23,解得a1=-1,∴数列{an}的通项公式为an=-3n+2.(2)由数列{an+bn}是首项为1,公比为2的等比数列,∴an+bn=2n-1,∴bn=2n-1-an=3n-2+2n-1,∴Sn=[1+4+7+…+(3n-2)]+(1+2+22+…+2n-1)=n(3n-1)2+2n-1.5.解(1)设等比数列{bn}的公比为q.由b1=1,b3=b2+2,可得q2-q-2=0.因为q>0,可得q=2,故bn=2n-1.所以,Tn=1-2n1-2=2n-1.设等差数列{an}的公差为d.由b4=a3+a5,可得a1+3d=4.由b5=a4+2a6,可得3a1+13d=16,从而a1=1,d=1,故an=n.所以,Sn=n(n+1)2.(2)由(1),有T1+T2+…+Tn=(21+22+…+2n)-n=2×(1-2n)1-2-n=2n+1-n-2.由Sn+(T1+T2+…+Tn)=an+4bn可得,n(n+1)2+2n+1-n-2=n+2n+1,整理得n2-3n-4=0,解得n=-1(舍),或n=4.所以,n的值为4.6.解(1)设等差数列{an}的公差为d,则an=a1+(n-1)d.因为a7=8,所以a1+6d=8.又a19=2a9,所以a1+18d=2(a1+8d),解得a1=2,d=1,所以{an}的通项公式为an=n+1.4\n(2)bn=1nan=1n(n+1)=1n-1n+1,所以Sn=1-12+12-13+…+1n-1n+1=nn+1.7.解(1)设{an}的公比为q,由题意知a1(1+q)=6,a12q=a1q2,又an>0,解得a1=2,q=2,所以an=2n.(2)由题意知S2n+1=(2n+1)(b1+b2n+1)2=(2n+1)bn+1,又S2n+1=bnbn+1,bn+1≠0,所以bn=2n+1.令cn=bnan,则cn=2n+12n,因此Tn=c1+c2+…+cn=32+522+723+…+2n-12n-1+2n+12n.又12Tn=322+523+724+…+2n-12n+2n+12n+1,两式相减得12Tn=32+12+122+…+12n-1-2n+12n+1,所以Tn=5-2n+52n.8.解(1)设数列{an}的公差为d,数列{bn}的公比为q,q>0,∵b1=-2a1=2,a3-b2=-1,S3-2b3=7,∴a1=-1,-1+2d-2q=-1,3×(-1)+3d-2×2q2=7,解得d=2,q=2.∴an=-1+2(n-1)=2n-3,bn=2n.(2)cn=(-1)n-1anbn=(-1)n-1(2n-3)2n,∴Tn=-12-122+323-524+…+(-1)n-2(2n-5)2n-1+(-1)n-1(2n-3)2n,12Tn=-122-123+324+…+(-1)n-2(2n-5)2n+(-1)n-1(2n-3)2n+1,∴32Tn=-12-12+122-123+…+(-1)n-1×12n-1+(-1)n-1(2n-3)2n+1=-12+-121--12n-11--12+(-1)n-1(2n-3)2n+1,∴Tn=-59+29-12n-1+(-1)n-1(2n-3)3×2n.4