第2課時數(shù)列求和之裂項相消法求和考點01裂項相消法之根式型例1已知數(shù)列{an}的前n項和為Sn,且Sn=n2,則1a1+a2+1a2+aA.4010-12 B.4051-12C.例2設(shè)公差不為0的等差數(shù)列an的首項為2，且a（1）求數(shù)列an（2）已知數(shù)列bn為正項數(shù)列，且bn2=an+43，設(shè)數(shù)列例3：設(shè)公差不為0的等差數(shù)列an的首項為1，且a（1）求數(shù)列an（2）已知數(shù)列bn為正項數(shù)列，且bn2=an+12，設(shè)數(shù)列考點02裂項相消法之等差型例4數(shù)列{an}是公差為2的等差數(shù)列,a5=11,則數(shù)列1anan+1A.200101 B.100603C.100609 例5已知Sn是數(shù)列an的前n項和，數(shù)列（1）求數(shù)列an（2）已知cn=?2nan，求數(shù)列考點03裂項相消法之指數(shù)型例6已知數(shù)列an，若a1=2（1）證明數(shù)列an+1（2）若bn=n(an+1)3n，且數(shù)列（3）若cn=2(an+1)ana例7已知正項數(shù)列an的首項為7，且an+12?3an+1=（1）求an和b（2）求數(shù)列an+bn的前（3）設(shè)cn=annn+1bn，Tn為數(shù)列c題型04裂項相消法之裂項相加型例8已知數(shù)列an的前n項和為Sn，且滿足nSn+1?(n+1)（1）)求數(shù)列an（2）若bn=(?1)n?2a例9已知數(shù)列an是公差大于2的等差數(shù)列，其前n項和為Sn，a2=5，且a1（1）求數(shù)列an（2）令bn=(?1)n+12an限時訓(xùn)練1.數(shù)列1(n+1)(n+3)的前n項和Sn=A.512-2n+52(n+2)(n+3)B.34-22.數(shù)列{an},{bn}滿足an·bn=1,an=n2+5n+6,則{bn}的前100項和S100= ()A.2551 B.100309C.12 3.數(shù)列{an}的通項公式為an=12n+1+2n-1,前n項和為Sn,若SA.51 B.40C.41 D.504.已知數(shù)列{an}的前n項和Sn=n2,記數(shù)列1anan+1的前n項和為Tn,則T2025A.40474048 B.20244049C.40484049 5.數(shù)列{an}滿足an=1+2+3+…+nn,則數(shù)列1anan+1的前A.2n+2 B.2nn+2 C.6.已知等差數(shù)列{an}的前n項和為Sn,a1=9,a2為整數(shù),且Sn≤S5恒成立,則數(shù)列2anan+1的前9項和TA.-29 B.29 C.-27 7.數(shù)列{an}的前n項和為Sn,且Sn=n2+2n,bn=2an+an+1(n∈N*),則數(shù)列{bn}的前n項和TA.2n+1-2n-1 B.2n+3-1C.2n-8.(多選題)已知等差數(shù)列{an}的前n項和為Sn,a2=4,S7=42,則下列結(jié)論正確的是 ()A.a5=4B.Sn=12n2+5C.ann為遞減數(shù)列D.19.已知數(shù)列{an},a1=1,對于任意正整數(shù)n,都滿足an+1-an=n-1,則1a3-1+1a4-10.在等差數(shù)列{bn}中,b1=1,已知數(shù)列{an}滿足bn=log3a2n-1,且12是a1,a3+21的等比中項.(1)求{bn}的通項公式;(2)求數(shù)列1bnbn+1的前11.(多選題)設(shè)數(shù)列{an}的前n項和為Sn,已知Sn=2an-1,則下列結(jié)論正確的是 ()A.S2=2B.數(shù)列{an}為等比數(shù)列C.an=2nD.若bn=1log2an12.已知數(shù)列{an}的通項公式為an=12n-1,數(shù)列{bn}的通項公式為bn=1n2+2n,將數(shù)列{an}與數(shù)列{bn}的公共項按從大到小的順序排列組成一個新數(shù)列{A.12 B.99199 C.99197 13.設(shè)數(shù)列{an}的通項公式為an=2n+1,數(shù)列2nanan+1的前m項和Tm=14.已知數(shù)列{an}滿足a1=4,且an+1=3an-2n+1.在數(shù)列{bn}中,b1=1,bn+1=bn+1n(1)證明{an-n}是等比數(shù)列,并求{an}的通項公式;(2)求{bn}的通項公式;(3)記數(shù)列{cn}滿足cn=nbn,n為奇數(shù),n(an-

參考答案1.A[解析]因為1(n+1)(n+3)=121n+1-1n2.B[解析]∵an·bn=1,∴bn=1n2+5n+6=1(n+2)(n+3)=1n+2-1n+3,∴S100=13-13.B[解析]因為an=12n+1+2n-1=12(2n+1-2n-1),所以Sn=a1+a2+a3+…+an=12(3-1+5-3+…+2n+1-2n-4.D[解析]因為Sn=n2①,所以當(dāng)n≥2時,Sn-1=(n-1)2②,①-②得an=Sn-Sn-1=n2-(n-1)2=2n-1(n≥2),當(dāng)n=1時,a1=S1=1滿足an=2n-1,所以an=2n-1,則1anan+1=1(2n-1)(2n+15.B[解析]因為an=1+2+3+…+nn=(1+n)n2n=1+n2,所以1anan+1=46.A[解析]設(shè)等差數(shù)列{an}的公差為d,由Sn≤S5,得S4≤S5,S6≤S5,則a5≥0,a6≤0,即9+4d≥0,9+5d≤0,解得-94≤d≤-95,又a2為整數(shù),所以d為整數(shù),所以d=-2,則an=11-2n,則2anan+17.D[解析]因為Sn=n2+2n,所以當(dāng)n≥2時,an=Sn-Sn-1=n2+2n-[(n-1)2+2(n-1)]=2n+1,又a1=S1=3滿足上式,所以an=2n+1,則bn=22n+1+2n+3=2n+3-2n+1,所以Tn=(5-3)+(7-5)+(9-7)+…+(8.BC[解析]由S7=7(a1+a7)2=7a4=42,解得a4=6,又a2=4,所以{an}的公差d=a4-a24-2=1,所以an=a2+(n-2)d=n+2.對于A,a5=7,A錯誤;對于B,a1=3,則Sn=n(3+n+2)2=12n2+52n,B正確;對于C,ann=1+2n,則an+1n+1-ann=1+2n+1-1-2n=-2n(n+1)<0,所以a9.20231012[解析]因為an+1-an=n-1,所以當(dāng)n≥2時,an-an-1=n-2,an-1-an-2=n-3,…,a3-a2=1,a2-a1=0,所以當(dāng)n≥2時,an=a1+(a2-a1)+(a3-a2)+…+(an-an-1)=1+0+1+2+…+n-2=1+(n-1)(n-2)2,又a所以當(dāng)n≥3時,1an-1=2(n-1)(n-2)=21n-2-10.解:(1)由b1=log3a1=1,得a1=3.因為12是a1,a3+21的等比中項,所以122=a1(a3+21),則a3=48-21=27,則b2=log3a3=log327=3.設(shè){bn}的公差為d,則d=b2-b1=2,所以bn=b1+(n-1)d=2n-1.(2)由(1)可知1bnbn+1=1(2n-1)(2n+1)=1211.BD[解析]當(dāng)n=1時,由Sn=2an-1,得S1=a1=2a1-1,解得a1=1,當(dāng)n≥2時,an=Sn-Sn-1=2an-1-(2an-1-1)=2an-2an-1,即an=2an-1(n≥2),所以數(shù)列{an}是以1為首項,2為公比的等比數(shù)列,則an=2n-1,a2=2,S2=3,故A,C錯誤,B正確;因為bn=1log2an+1log2an+2=1n(n+1)=1n-1n+1,所以數(shù)列{12.B[解析]數(shù)列{2n-1}是由所有的正奇數(shù)構(gòu)成的數(shù)列.n2+2n=(n+1)2-1,當(dāng)n=2k-1(k∈N*)時,(n+1)2-1=4k2-1為奇數(shù);當(dāng)n=2k(k∈N*)時,(n+1)2-1=(2k+1)2-1=4k(k+1)為偶數(shù).所以cn=14n2-1=1(2n-1)(2n+1)=1212n-1-13.8[解析]由題意得2nanan+1=2n(2n+1)(2n+1+1)=12n+1-12n+1+1,則Tm=13-1514.解:(1)由an+1=3an-2n+1,得an+1-(n+1)=3an-3n=3(an-n),又a1-1=3≠0,所以an+1-(n+1)an-n=3,所以{an-n}是首項為3,公比為3的等比數(shù)列.故an(2)由題意可得bn+1-bn=1n(n+1)所以當(dāng)n≥2時,bn=b1+(b2-b1)+(b3-b2)+…+(bn-bn-1)=1+1-12+12-13+…+1n-1-1n又b1=1也滿足上式,所以bn=2n(3)由(1)(2)知cn=2n-1,n為奇數(shù),n·3n,n為偶數(shù),在{cn}的前2n項中,設(shè)T奇T偶=c2+