高考數(shù)學二輪總復(fù)習專題能力訓(xùn)練12數(shù)列的通項與求和理

專題能力訓(xùn)練12數(shù)列的通項與求和能力突破訓(xùn)練1.已知數(shù)列{an}是等差數(shù)列,前n項和為Sn,且滿足a1+2a2=S5,則下列結(jié)論錯誤的是()A.S9=0 B.S5最小 C.S3=S6 D.a5=02.在數(shù)列{an}中,an+1=2an1,a3=2,設(shè)其前n項和為Sn,則S6=()A.874 B.634 C.15 D3.已知數(shù)列{an}的前n項和為Sn,a1=1,a2=2,且對于任意n>1,n∈N*,滿足Sn+1+Sn1=2(Sn+1),則S10的值為()A.90 B.91 C.96 D.1004.(2022廣東廣州三模)已知數(shù)列{an}滿足a1a2…an=22an,bn=an(n+1)2A.10112024 B.10212024 C.51110125.已知數(shù)列{an},構(gòu)造一個新數(shù)列a1,a2a1,a3a2,…,anan1,…,此數(shù)列是首項為1,公比為13的等比數(shù)列,則數(shù)列{an}的通項公式為(A.an=32-32×1B.an=32+32×C.an=1D.an=1,n∈N*6.(2022廣西四市4月質(zhì)檢)已知數(shù)列{an}滿足a1+a22+a33+…+an-7.已知數(shù)列{an}的前n項和為Sn,a1=a2=1,當n∈N*時,an+2=(1)n(an+n)n,則S20=.

8.已知Sn是等差數(shù)列{an}的前n項和,若a1=1017,S10141014-S10081008=6,則9.(2022新高考Ⅰ,17)記Sn為數(shù)列{an}的前n項和,已知a1=1,Snan是公差為(1)求{an}的通項公式;(2)證明:1a1+1a2+10.已知數(shù)列{an}是公差為2的等差數(shù)列,它的前n項和為Sn,且a1,a3,a7成等比數(shù)列.(1)求數(shù)列{an}的通項公式;(2)設(shè)數(shù)列{bn}滿足bn=Sn-n23n+1,求數(shù)列{b11.已知公差不為0的等差數(shù)列{an}滿足a3=5,且a1,a2,a5成等比數(shù)列.(1)求{an}的通項公式;(2)設(shè)bn=13n-1anan+1思維提升訓(xùn)練12.在數(shù)列{an}中,如果?n∈N*,都有anan+1·an+2=k(k為常數(shù)),那么數(shù)列{an}叫做等積數(shù)列,k叫做這個數(shù)列的公積.已知數(shù)列{an}是等積數(shù)列,且a1=1,a2=2,公積為8,則a1+a2+…+a1020=()A.2377 B.2380 C.2378 D.238113.(2022廣西師大附屬外國語學校模擬)設(shè)等差數(shù)列{an}的前n項和為Sn,等比數(shù)列{bn}的前n項和Tn=3n1k,數(shù)列{cn}滿足c1=2,c2=3,c3=b3,且cn+1=cn+an,則下列幾個結(jié)論中,所有正確結(jié)論的序號為.

①k=3;②an2+bn2>an+12+bn+12;14.(2022山東濰坊二模)已知正項數(shù)列{an}的前n項和為Sn,且an2+2an=4Sn,數(shù)列{bn}滿足bn=(2(1)求數(shù)列{bn}的前n項和Bn,并證明Bn+1,Bn,Bn+2成等差數(shù)列;(2)設(shè)cn=(1)nan+bn,求數(shù)列{cn}的前n項和Tn.15.已知等比數(shù)列{an}的前n項和為Sn(Sn≠0),若S1,S2,S3成等差數(shù)列,且a1a2=a3.(1)求數(shù)列{an}的通項公式;(2)設(shè)bn=-3an(an+1)(a16.已知{an}為等差數(shù)列,{bn}為等比數(shù)列,a1=b1=1,a5=5(a4a3),b5=4(b4b3).(1)求{an}和{bn}的通項公式;(2)記{an}的前n項和為Sn,求證:SnSn+2<Sn+12(n∈(3)對任意的正整數(shù)n,設(shè)cn=(3an-2)b

專題能力訓(xùn)練12數(shù)列的通項與求和能力突破訓(xùn)練1.B解析由題設(shè)可得3a1+2d=5a1+10d?2a1+8d=0,即a5=0,所以D中結(jié)論正確.由等差數(shù)列的性質(zhì)可得a1+a9=2a5=0,則S9=9(a1+a9)S3S6=3a1+3d6a115d=3(a1+4d)=3a5=0,所以C中結(jié)論正確.B中結(jié)論是錯誤的.故選B.2.A解析由an+1=2an1,得an+11=2(an1),又a3=2,所以2a21=2,所以a2=32,所以2a11=32,所以a1=所以{an1}是等比數(shù)列,首項為14則an1=14·2n1=2n3,即an=1+2n3,S6=63.B解析∵Sn+1+Sn1=2(Sn+1),∴當n≥2時,Sn+1Sn=SnSn1+2,∴當n≥2時,an+1an=2.∴當n≥2時,數(shù)列{an}是等差數(shù)列,公差為2.又a1=1,a2=2,∴S10=1+9×2+9×82×4.A解析當n=1時,有a1=22a1,所以a1=23因為a1a2…an=22an, ①所以當n≥2時,a1a2…an1=22an1. ②①÷②得an=1-an1-an-1(n≥2),所以an=1an所以anan1=1an+anan1an1=(1an)(1an1)(n≥2),所以an-an又11-所以11-an是一個首項為3,公差為1的等差數(shù)列,所以11-所以bn=1(所以數(shù)列{bn}的前2022項和為12-13+15.A解析因為數(shù)列a1,a2a1,a3a2,…,anan1,…是首項為1,公比為13所以anan1=13n-1,n≥2an=a1+(a2a1)+(a3a2)+…+(anan1)=1+13+13又當n=1時,a1=1也適合上式,所以an=32-32×16.n解析因為a1+a22+a33+所以當n=1時,a1=1,當n≥2時,a1+a22+a33+①②可得,ann=n(n1)=1,即an當n=1時,an=n也成立,綜上可知,an=n.7.80解析當n為奇數(shù)時,an+2+an=2n;當n為偶數(shù)時,an+2=an=a2=1.所以S20=(a1+a3+a5+a7+…+a17+a19)+(a2+a4+a6+a8+…+a18+a20)=2×(1+5+9+13+17)+10=80.8.1017解析∵Sn是等差數(shù)列{an}的前n項和,∴Snn∵S10141014-S10081008=6,∵a1=1017,∴S11∴Snn=1017+(n1)×1=∴S1017=(1018+1017)×1017=1017.9.(1)解方法一:∵Snan是以S1a1=1為首項,以13為公差的等差數(shù)列,∴Sn=n+23an.當n≥2時,Sn1=n+13an1.①②得an=SnSn1=n+23ann+13∴n+13an1=n-∴an=anan-1·an-1an-又a1=1,∴an=(n+1)n2×1又當n=1時,a1=1也符合上式,∴an=n方法二:∵Snan是以S1a1=1為首項,以13為公差的等差數(shù)列,∴Sn=n+23an.當n≥2時,Sn1=n+13an1.①②得an=SnSn1=n+23ann+13∴n+13an1=n∴an設(shè)ann(n+1)=bn,則∴{bn}為常數(shù)列,且b1=a1∴ann(n+1)=bn=(2)證明由(1)知,1an=2∴1a1+1a2+…+1an=2112+12-110.解(1)因為數(shù)列{an}是公差為2的等差數(shù)列,且a1,a3,a7成等比數(shù)列,所以a32=a1·a7,即(a1+4)2=a1(a1+12),解得a1=所以an=4+2(n1)=2n+2.(2)由(1)得Sn=n(4+2n+2)2=n2+3n所以Tn=13+232+13Tn=132+①②,得23Tn=13+132+1311.解(1)設(shè)等差數(shù)列{an}的公差為d(d≠0).由a1,a2,a5成等比數(shù)列,得a22=a1a即(a3d)2=(a32d)(a3+2d),所以(5d)2=(52d)(5+2d),又d≠0,所以d=2.所以a1=a32d=1,所以an=2n1.(2)由(1)得bn=13所以Tn=131-13n1-13思維提升訓(xùn)練12.B解析由題意,得a1·a2·a3=8,即1×2a3=8,于是a3=4.同理可求a4=1,a5=2,a6=4,….∴{an}是以3為周期的數(shù)列,∴a1=a4=a7=a10=1,a2=a5=a8=a11=2,a3=a6=a9=a12=4.∴a1+a2+…+a1020=340(a1+a2+a3)=340×7=2380.13.③④解析對于①,當n=1時,b1=T1=1k,當n≥2時,bn=TnTn1=(3n1k)(3n2k)=2·3n2,因為{bn}是等比數(shù)列,所以b1=1k=2·312,所以k=13,①對于②,因為c3=b3=6,a1=c2c1=1,a2=c3c2=3.又{an}是等差數(shù)列,所以公差d=a2a1=2,所以an=1+2(n1)=2n1.所以Sn=n(1+2n設(shè)f(n)=an2+bn2=(2n1)2+4·9n2,則f(n+1)=(2n+1)2+所以f(n+1)f(n)=8n+32·9n2>0,即f(n)<f(n+1),②錯誤;對于③,SnSn+2Sn+12=n2(n+2)2(n+1)4=[n(n+2)(n+1)2]·[n(n+2)+(n+1)2]=(2n2+4n+1)<對于④,因為cn+1=cn+2n1,所以當n≥2時,cn=c1+(c2c1)+(c3c2)+…+(cncn1)=2+1+3+…+(2n3)=2+(1+2n-3)(n-當n=1時,c1=2滿足cn=(n1)2+2,所以cn=(n1)2+2.所以1故1c1+1c2+…+1cn≥11-12+12-1314.解(1)an2+2an=4Snan>0,當n=1時,a12+2a1=4a1,∴a1=2或a1當n≥2時,an-12+2an1=4S①②得an2-an-12+2an2∴(anan1)(an+an1)=2(an+an1),∵an>0,∴anan1=2(n≥2),∴{an}是以2為首項,2為公差的等差數(shù)列,∴an=2n,∴bn=(2)n.∴數(shù)列{bn}是首項為2,公比為2的等比數(shù)列,∴Bn=-2[1-(-2)∴Bn+2+Bn+1=43+(1)n+2·2n+33+(1)n+1·2n+23=43+(-1)n+1·2n+2(2)cn=(1)n·2n+(2)n=(2)n+2(1)n·n,當n為偶數(shù)時,Tn=(2)1+(2)2+…+(2)n+2[1+23+4+…(n1)+n]=-2×[1-(-2)n]1-(-2)+2×[(1+2)+(3+4)+…當n為奇數(shù)時,Tn=(2)1+(2)2+…+(2)n+2[1+23+4+…+(n1)n]=-2×[1-(-2)n]1-(-2)+2×[1+(23)+(45)+…+(n1綜上可知,Tn=215.解(1)設(shè)數(shù)列{an}的公比為q,依題意,得S1+(S3)=2S2,所以(a2+a3)=2(a1+a2),得a1(q+q2)=2a1(1+q),且a1≠0,所以q2+3q+2=0,解得q=1或q=2,因為Sn≠0,所以q=2.又因為a1a2=a3,所以a12q=a1q2,所以a1=q,所以an=(2)(2)由(1)得bn=-3所以Tn=1(-2)1+1-1(-2)2+1+1(-16.(1)解設(shè)等差數(shù)列{an}的公差為d,等比數(shù)列{bn}的公比為q.由a1=1,a5=5(a4a3),可得d=1,從而{an}的通項公式為an=n.由b1=1,b5=4(b4b3),又q≠0,可得q24q+4=0,解