2019-2020年高考數(shù)學(xué)二輪復(fù)習(xí) 專題能力訓(xùn)練11 數(shù)列求和及綜合應(yīng)用 文一、選擇題1.在等差數(shù)列an中,a1+a5=8,a8=19,則其前10項(xiàng)的和S10等于()A.100B.115C.95D.852.在等比數(shù)列an中,a1=2,前n項(xiàng)的和為Sn,若數(shù)列an+1也是等比數(shù)列,則Sn等于()A.2nB.3nC.3n-1D.2n+1-23.已知Sn是非零數(shù)列an的前n項(xiàng)和,且Sn=2an-1,則Sxx等于()A.1-2xxB.2xx-1C.2xx-1D.2xx4.已知等比數(shù)列an的前n項(xiàng)和為Sn,若a2a3=2a1,且a4與2a7的等差中項(xiàng)為,則S5等于()A.35B.33C.31D.295.等差數(shù)列an的前n項(xiàng)和為Sn,已知a5+a7=4,a6+a8=-2,則當(dāng)Sn取最大值時(shí)n的值是()A.5B.6C.7D.86.若向量an=(cos2n,sin n),bn=(1,2sin n)(nN*),則數(shù)列anbn+2n的前n項(xiàng)和Sn等于()A.n2B.n2+2nC.2n2+4nD.n2+n二、填空題7.等比數(shù)列an的前n項(xiàng)和為Sn,且4a1,2a2,a3成等差數(shù)列,若a1=1,則S4=.8.已知數(shù)列an滿足a1=,且對(duì)任意的正整數(shù)m,n都有am+n=aman,則數(shù)列an的前n項(xiàng)和Sn=.9.對(duì)于數(shù)列an,定義數(shù)列an+1-an為數(shù)列an的“差數(shù)列”,若a1=2,an的“差數(shù)列”的通項(xiàng)為2n,則數(shù)列an的前n項(xiàng)和Sn=.三、解答題10.(xx四川成都三診)已知正項(xiàng)等比數(shù)列an滿足a2=,a4=,nN*.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)數(shù)列bn滿足bn=log3anlog3an+1,求數(shù)列的前n項(xiàng)和Tn.11.已知函數(shù)f(x)=,數(shù)列an滿足a1=1,an+1=f,nN*.(1)求數(shù)列an的通項(xiàng)公式;(2)令bn=(n2),b1=3,Sn=b1+b2+bn,若Sn<對(duì)一切nN*成立,求最小正整數(shù)m.12.設(shè)數(shù)列bn的前n項(xiàng)和為Sn,且bn=2-2Sn;數(shù)列an為等差數(shù)列,且a5=14,a7=20.(1)求數(shù)列bn的通項(xiàng)公式;(2)若cn=anbn(nN*),求數(shù)列cn的前n項(xiàng)和Tn.專題能力訓(xùn)練11數(shù)列求和及綜合應(yīng)用1.B解析:由a1+a5=8,得a3=4,S10=5(a3+a8)=523=115.故選B.2.A解析:設(shè)等比數(shù)列an的公比為q,則(a2+1)2=(a1+1)(a3+1),(2q+1)2=3(2q2+1),整理得q2-2q+1=0.即q=1,an=2,Sn=2n.選A.3.B解析:Sn=2an-1,Sn-1=2an-1-1(n2),兩式相減得an=2an-2an-1,即an=2an-1,數(shù)列an是公比為2的等比數(shù)列,由S1=2a1-1,得a1=1,Sxx=2xx-1.選B.4.C解析:設(shè)等比數(shù)列an的公比為q,由a2a3=2a1,得a1q3=2,即a4=2.由題意a4+2a7=,a7=.則q3=,q=.又a1q3=2,a1=16.S5=32-1=31.故選C.5.B解析:由a5+a7=4,a6+a8=-2,兩式相減得2d=-6,d=-3.a5+a7=4,2a6=4,即a6=2.由a6=a1+5d,得a1=17.an=a1+(n-1)(-3)=20-3n,令an>0,得n<,前6項(xiàng)和最大.故選B.6.B解析:anbn+2n=cos2n+2sin2n+2n=(1-2sin2n)+2sin2n+2n=2n+1,則數(shù)列anbn+2n是等差數(shù)列,Sn=n2+2n.故選B.7.15解析:設(shè)等比數(shù)列an的公比為q,由題意得4a1+a3=4a2,即q2-4q+4=0,q=2.故S4=15.8.2-解析:令m=1,則an+1=a1an,數(shù)列an是以a1=為首項(xiàng),為公比的等比數(shù)列,Sn=2-.9.2n+1-2解析:an+1-an=2n,an=(an-an-1)+(an-1-an-2)+(a2-a1)+a1=2n-1+2n-2+22+2+2=+2=2n,Sn=2n+1-2.10.解:(1)設(shè)公比為q.=q2,q=或q=-.又?jǐn)?shù)列an為正項(xiàng)等比數(shù)列,q=.又a2=,a1=.an=,nN*.(2)bn=log3anlog3an+1,nN*,bn=n(n+1),nN*.Tn=1-+=1-.11.解:(1)an+1=f=an+,an是以1為首項(xiàng),為公差的等差數(shù)列.an=1+(n-1)n+.(2)當(dāng)n2時(shí),bn=,又b1=3=,Sn=b1+b2+bn=.Sn<對(duì)一切nN*成立,.又遞增,且,即mxx.最小正整數(shù)m=xx.12.解:(1)由bn=2-2Sn,令n=1,則b1=2-2S1,b1=.當(dāng)n2時(shí),由bn=2-2Sn,可得bn-bn-1=-2(Sn-Sn-1)=-2bn,即,所以bn是以b1=為首項(xiàng),為公比的等比數(shù)列,則bn=.(2)數(shù)列an為等差數(shù)列,公差d=(a7-a5)=3,可得an=3n-1,從而cn=anbn=2(3n-1),Tn=2,Tn=2,Tn=2,Tn=.