考点规范练31数列求和基础巩固1.数列1,3,5,7,…,(2n-1)+,…的前n项和Sn的值等于()A.n2+1-B.2n2-n+1-C.n2+1-D.n2-n+1-2.已知数列{an}满足a1=1,且对任意的n∈N*都有an+1=a1+an+n,则的前100项和为()A.B.C.D.3.已知函数f(n)=n2cos(nπ),且an=f(n)+f(n+1),则a1+a2+a3+…+a100=()A.0B.-100C.100D.102004.已知函数f(x)=xa的图象过点(4,2),令an=,n∈N*.记数列{an}的前n项和为Sn,则S2018等于()A.-1B.+1C.-1D.+15.已知数列{an}满足an+1+(-1)nan=2n-1,则{an}的前60项和为()A.3690B.3660C.1845D.18306.3·2-1+4·2-2+5·2-3+…+(n+2)·2-n=.7.已知数列{an}满足:a3=,an-an+1=2anan+1,则数列{anan+1}前10项的和为.8.已知{an}是等差数列,{bn}是等比数列,且b2=3,b3=9,a1=b1,a14=b4.(1)求{an}的通项公式;(2)设cn=an+bn,求数列{cn}的前n项和.9.设等差数列{an}的公差为d,前n项和为Sn,等比数列{bn}的公比为q,已知b1=a1,b2=2,q=d,S10=100.1(1)求数列{an},{bn}的通项公式;(2)当d>1时,记cn=,求数列{cn}的前n项和Tn.10.已知Sn为数列{an}的前n项和,an>0,+2an=4Sn+3.(1)求{an}的通项公式;(2)设bn=,求数列{bn}的前n项和.11.已知各项均为正数的数列{an}的前n项和为Sn,满足=2Sn+n+4,a2-1,a3,a7恰为等比数列{bn}的前3项.(1)求数列{an},{bn}的通项公式;(2)若cn=(-1)nlog2bn-,求数列{cn}的前n项和Tn.2能力提升12.(2017山东烟台模拟)已知在数列{an}中,a1=1,且an+1=,若bn=anan+1,则数列{bn}的前n项和Sn为()A.B.C.D.13.(2017福建龙岩一模)已知Sn为数列{an}的前n项和,对n∈N*都有Sn=1-an,若bn=log2an,则+…+=.14.已知首项为的等比数列{an}不是递减数列,其前n项和为Sn(n∈N*),且S3+a3,S5+a5,S4+a4成等差数列.(1)求数列{an}的通项公式;(2)设bn=(-1)n+1·n(n∈N*),求数列{an·bn}的前n项和Tn.315.已知数列{an}的前n项和为Sn,且a1=0,对任意n∈N*,都有nan+1=Sn+n(n+1).(1)求数列{an}的通项公式;(2)若数列{bn}满足an+log2n=log2bn,求数列{bn}的前n项和Tn.高考预测16.已知数列{an}的前n项和为Sn,且a1=2,Sn=2an+k,等差数列{bn}的前n项和为Tn,且Tn=n2.(1)求k和Sn;4(2)若cn=an·bn,求数列{cn}的前n项和Mn.答案：1.A解析:该数列的通项公式为an=(2n-1)+,则Sn=[1+3+5+…+(2n-1)]+=n2+1-.2.D解析: an+1=a1+an+n,∴an+1-an=1+n.∴an-an-1=n(n≥2).∴an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=n+(n-1)+…+2+1=.∴=2.∴的前100项和为2=2.故选D.3.B解析: f(n)=n2cos(nπ)==(-1)n·n2,∴an=f(n)+f(n+1)=(-1)n·n2+(-1)n+1·(n+1)2=(-1)n[n2-(n+1)2]=(-1)n+1·(2n+1).∴a1+a2+a3+…+a100=3+(-5)+7+(-9)+…+199+(-201)=50×(-2)=-100.故选B.4.C解析:由f(4)=2,可得4a=2,解得a=,则f(x)=.∴an=,S2018=a1+a2+a3+…+a2018=()+()+()+…+()=-1.5.D解析: an+1+(-1)nan=2n-1,∴当n=2k(k∈N*)时,a2k+1+a2k=4k-1,①当n=2k+1(k∈N*)时,a2k+2-a2k+1=4k+1,②①+②得:a2k+a2k+2=8k.则a2+a4+a6+a8+…+a60=(a2+a4)+(a6+a8)+…+(a58+a60)=8(1+3+…+29)=8×=1800.由②得a2k+1=a2k+2-(4k+1),∴a1+a3+a5+…+a59=a2+a4+…+a60-[4×(0+1+2+…+29)+30]=1800-=30,∴a1+a2+…+a60=1800+30=1830.6.4-解析:设Sn=3·2-1+4·2-2+5·2-3+…+(n+2)·2-n,①则Sn=3·2-2+4·2-3+…+(n+1)2-n+(n+2)2-n-1.②①-②,得Sn=3·2-1+2-2+2-3+…+2-n-(n+2)·2-n-1=2·2-1+2-1+2-2+2-3+…+2-n-(n+2)·2-n-1=1+-(n+2)·2-n-1=2-(n+4)·2-n-1.故Sn=4-.7.解析: an-an+1=2anan+1,∴=2,即=2.∴数列是以2为公差的等差数列. =5,∴=5+2(n-3)=2n-1.∴an=.∴anan+1=.∴数列{anan+1}前10项的和为=.8.解:(1)因为等比数列{bn}的公比q==3,5所以b1==1,b4=b3q=27.设等差数列{an}的公差为d.因为a1=b1=1,a14=b4=27,所以1+13d=27,即d=2.所以an=2n-1.(2)由(1)知,an=2n-1,bn=3n-1.因此cn=an+bn=2n-1+3n-1.从而数列{cn}的前n项和Sn=1+3+…+(2n-1)+1+3+…+3n-1==n2+.9.解:(1)由题意,有解得故(2)由d>1,知an=2n-1,bn=2n-1,故cn=,于是Tn=1++…+,①Tn=+…+.②①-②可得Tn=2++…+=3-,故Tn=6-.10.解:(1)由+2an=4Sn+3,可知+2an+1=4Sn+1+3.两式相减可得+2(an+1-an)=4an+1,即2(an+1+an)==(an+1+an)·(an+1-an).由于an>0,...
