2019-2020年高考数学二轮复习 专题能力训练11 数列求和及综合应用 文.doc

2019-2020年高考数学二轮复习 专题能力训练11 数列求和及综合应用 文一、选择题1.在等差数列an中,a1+a5=8,a8=19,则其前10项的和S10等于()A.100B.115C.95D.852.在等比数列an中,a1=2,前n项的和为Sn,若数列an+1也是等比数列,则Sn等于()A.2nB.3nC.3n-1D.2n+1-23.已知Sn是非零数列an的前n项和,且Sn=2an-1,则Sxx等于()A.1-2xxB.2xx-1C.2xx-1D.2xx4.已知等比数列an的前n项和为Sn,若a2a3=2a1,且a4与2a7的等差中项为,则S5等于()A.35B.33C.31D.295.等差数列an的前n项和为Sn,已知a5+a7=4,a6+a8=-2,则当Sn取最大值时n的值是()A.5B.6C.7D.86.若向量an=(cos2n,sin n),bn=(1,2sin n)(nN*),则数列anbn+2n的前n项和Sn等于()A.n2B.n2+2nC.2n2+4nD.n2+n二、填空题7.等比数列an的前n项和为Sn,且4a1,2a2,a3成等差数列,若a1=1,则S4=.8.已知数列an满足a1=,且对任意的正整数m,n都有am+n=aman,则数列an的前n项和Sn=.9.对于数列an,定义数列an+1-an为数列an的“差数列”,若a1=2,an的“差数列”的通项为2n,则数列an的前n项和Sn=.三、解答题10.(xx四川成都三诊)已知正项等比数列an满足a2=,a4=,nN*.(1)求数列an的通项公式;(2)设数列bn满足bn=log3anlog3an+1,求数列的前n项和Tn.11.已知函数f(x)=,数列an满足a1=1,an+1=f,nN*.(1)求数列an的通项公式;(2)令bn=(n2),b1=3,Sn=b1+b2+bn,若Sn0,得n,前6项和最大.故选B.6.B解析:anbn+2n=cos2n+2sin2n+2n=(1-2sin2n)+2sin2n+2n=2n+1,则数列anbn+2n是等差数列,Sn=n2+2n.故选B.7.15解析:设等比数列an的公比为q,由题意得4a1+a3=4a2,即q2-4q+4=0,q=2.故S4=15.8.2-解析:令m=1,则an+1=a1an,数列an是以a1=为首项,为公比的等比数列,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)设公比为q.=q2,q=或q=-.又数列an为正项等比数列,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为首项,为公差的等差数列.an=1+(n-1)n+.(2)当n2时,bn=,又b1=3=,Sn=b1+b2+bn=.Sn对一切nN*成立,.又递增,且,即mxx.最小正整数m=xx.12.解:(1)由bn=2-2Sn,令n=1,则b1=2-2S1,b1=.当n2时,由bn=2-2Sn,可得bn-bn-1=-2(Sn-Sn-1)=-2bn,即,所以bn是以b1=为首项,为公比的等比数列,则bn=.(2)数列an为等差数列,公差d=(a7-a5)=3,可得an=3n-1,从而cn=anbn=2(3n-1),Tn=2,Tn=2,Tn=2,Tn=.