专题能力训练11数列求和及综合应用一、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S2014等于()A.1-22014B.22014-1C.22015-1D.220134.已知等比数列{an}的前n项和为Sn,若a2·a3=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=(cos2nθ,sinnθ),bn=(1,2sinnθ)(n∈N*),则数列{an·bn+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=am·an,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.(2014四川成都三诊)已知正项等比数列{an}满足a2=,a4=,n∈N*.(1)求数列{an}的通项公式;(2)设数列{bn}满足bn=log3an·log3an+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,n∈N*.(1)求数列{an}的通项公式;(2)令bn=(n≥2),b1=3,Sn=b1+b2+…+bn,若Sn<对一切n∈N*成立,求最小正整数m.12.设数列{bn}的前n项和为Sn,且bn=2-2Sn;数列{an}为等差数列,且a5=14,a7=20.(1)求数列{bn}的通项公式;(2)若cn=an·bn(n∈N*),求数列{cn}的前n项和Tn.答案与解析专题能力训练11数列求和及综合应用1.B解析:由a1+a5=8,得a3=4,∴S10==5(a3+a8)=5×23=115.故选B.2.A解析:设等比数列{an}的公比为q,则(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.选A.3.B解析:∵Sn=2an-1,∴Sn-1=2an-1-1(n≥2),两式相减得an=2an-2an-1,即an=2an-1,∴数列{an}是公比为2的等比数列,由S1=2a1-1,得a1=1,∴S2014==22014-1.选B.4.C解析:设等比数列{an}的公比为q,由a2·a3=2a1,得a1q3=2,即a4=2.由题意a4+2a7=,∴a7=.则q3=,∴q=.又a1q3=2,∴a1=16.∴S5==32-1=31.故选C.5.B解析:由a5+a7=4,a6+a8=-2,两式相减得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项和最大.故选B.6.B解析:an·bn+2n=cos2nθ+2sin2nθ+2n=(1-2sin2nθ)+2sin2nθ+2n=2n+1,则数列{an·bn+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=a1·an,∴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=,n∈N*.(2)∵bn=log3an·log3an+1,n∈N*,∴bn=n(n+1),n∈N*.∴.∴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n≥2时,bn==,又b1=3=,∴Sn=b1+b2+…+bn==.∵Sn<对一切n∈N*成立,∴.又递增,且,∴,即m≥2014.∴最小正整数m=2014.12.解:(1)由bn=2-2Sn,令n=1,则b1=2-2S1,∴b1=.当n≥2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=an·bn=2(3n-1)·,∴Tn=2,Tn=2,∴Tn=2,∴Tn=.
