《全品高考復習方案》第2課時裂項相消法求和1.A[解析]因為1(n+1)(n+3)=121n+1-1n+3,2.B[解析]∵an·bn=1,∴bn=1n2+5n+6=1(n+2)(n+3)=1n+2-1n+3,∴S100=13-143.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-1)=4.D[解析]因為Sn=n2①,所以當n≥2時,Sn-1=(n-1)2②,①-②得an=Sn-Sn-1=n2-(n-1)2=2n-1(n≥2),當n=1時,a1=S1=1滿足an=2n-1,所以an=2n-1,則1anan+1=1(2n-1)(2n+1)=5.B[解析]因為an=1+2+3+…+nn=(1+n)n2n=1+n2,所以1anan+1=4(n6.A[解析]設等差數(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+1=27.D[解析]因為Sn=n2+2n,所以當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)+…+(28.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ann為遞減數(shù)列,C正確;對于D,1a9.20231012[解析]因為an+1-an=n-1,所以當n≥2時,an-an-1=n-2,an-1-an-2=n-3,…,a3-a2=1,a2-a1=0,所以當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a1=1滿足上式,所以an=1+(n-1)(n-2)2,所以當n≥3時,1an-110.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{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則Sn=1211-13+121311.BD[解析]當n=1時,由Sn=2an-1,得S1=a1=2a1-1,解得a1=1,當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{bn}的前10項和為112.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n=2k-1(k∈N*)時,(n+1)2-1=4k2-1為奇數(shù);當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-1-1213.8[解析]由題意得2nanan+1=2n(2n+1)(2n+1+1)=12n+1-12n+1+1,則Tm=13-15+15-14.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{故an-n=3n,則an=3n+n.(2)由題意可得bn+1-bn=1n(n+1)所以當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=2在{cn}的前2n項中,設T奇=c1+c3+c5+…+c2n-1,T偶=c2+c4+c6+…+c2n.則T奇=1+5+9+…+4n-3=n+n(n-1)2×T偶=2×32+4×34+6×36+…+2n×32n=2×9+4×92+6×93+…+2n×9n