考点规范练30数列求和考点规范练B册第20页基础巩固组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-答案: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.(2015云南曲靖一模)+…+的值为()A.B.C.D.导学号〚32470776〛答案:C解析:∵,∴+…+=+…+==.3.已知数列{an}:,…,+…+,…,若bn=,那么数列{bn}的前n项和Sn等于()A.B.C.D.导学号〚32470777〛答案:B解析:易得an=,∴bn==4.∴Sn=4+…+=4.4.已知函数f(x)=xa的图像过点(4,2),令an=,n∈N+.记数列{an}的前n项和为Sn,则S2016等于()A.-1B.+1C.-1D.+1答案:C解析:由f(4)=2,可得4a=2,解得a=,则f(x)=.∴an=,S2016=a1+a2+a3+…+a2016=()+()+()+…+()=-1.5.数列{an}满足an+1+(-1)nan=2n-1,则{an}的前60项和为()A.3690B.3660C.1845D.1830答案: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=1830.6.3·2-1+4·2-2+5·2-3+…+(n+2)·2-n=.答案: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.已知在等比数列{an}中,a1=3,a4=81,若数列{bn}满足bn=log3an,则数列的前n项和Sn=.导学号〚32470778〛答案:解析:设等比数列{an}的公比为q,则=q3=27,解得q=3.∴an=a1qn-1=3×3n-1=3n,故bn=log3an=n,∴,则数列的前n项和为1-+…+=1-.8.(2015长春模拟)已知等差数列{an}满足:a5=9,a2+a6=14.(1)求{an}的通项公式;(2)若bn=an+(q>0),求数列{bn}的前n项和Sn.解:(1)∵在等差数列{an}中,a5=9,a2+a6=2a4=14,∴a4=7,其公差d=a5-a4=2,∴an=a4+(n-4)d=7+2(n-4)=2n-1.(2)∵bn=an+(q>0),∴Sn=b1+b2+…+bn=(a1+a2+…+an)+(+…+)=[1+3+5+…+(2n-1)]+(q1+q3+…+q2n-1)=n2+(q1+q3+…+q2n-1).若q=1,Sn=n2+n;若q≠1,Sn=n2+.9.(2015湖北,文19)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)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.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-.导学号〚32470779〛10.(2015山东,文19)已知数列{an}是首项为正数的等差数列,数列的前n项和为.(1)求数列{an}的通项公式;(2)设bn=(an+1)·,求数列{bn}的前n项和Tn.解:(1)设数列{an}的公差为d.令n=1,得,所以a1a2=3.令n=2,得,所以a2a3=15.解得a1=1,d=2,所以an=2n-1.(2)由(1)知bn=(an+1)·=2n·22n-1=n·4n,所以Tn=1·41+2·42+…+n·4n,所以4Tn=1·42+2·43+…+n·4n+1,两式相减,得-3Tn=41+42+…+4n-n·4n+1=-n·4n+1=×4n+1-.所以Tn=×4n+1+.11.在数列{an}中,a1=1,当n≥2时,其前n项和Sn满足=an.(1)求Sn的表达式;(2)设bn=,求数列{bn}的前n项和Tn.解:(1)∵=an,又an=Sn-Sn-1(n≥2),∴=(Sn-Sn-1),即2Sn-1Sn=Sn-1-Sn.①由题意得Sn-1·Sn≠0,①式两边同除以Sn-1·Sn,得=2,∴数列是首项为=1,公差为2的等差数列.∴=1+2(n-1)=2n-1,∴Sn=.(2)∵bn==,∴Tn=b1+b2+…+bn=+…+=.能力提升组12.已知正项数列{an},{bn}满足a1=3,a2=6,{bn}是等差数列,且对任意正整数n,都有bn,,bn+1成等比数列.(1)求数列{bn}的通项公式;(2)设Sn=+…+,试比较2Sn与2-的大小.解:(1)对任意正整数n,都有bn,,bn+1成等比数列,且{an},{bn}都为正项数列,∴an=bnbn+1.∴a1=b1b2=3,a2=b2b3=6.又{bn}是等差数列,∴b1+b3=2b2,解得b1=,b2=.∴bn=(n+1).(2)由(1)可得an=bnbn+1=,则=2,∴Sn=2+…+=1-.∴2Sn=2-.又2-=2-,∴2Sn-.∴当n=1,2时,2Sn<2-;当n≥3时,2Sn>2-.导学号〚32470780〛13.已知数列{an}是等差数列,{bn}是等比数列,其中a1=b1=1,a2≠b2,且b2为a1,a2的等差中项,a2为b2,b3的等差中项.(1)求数列{an}与{bn}的通项公式;(2)记cn=(a1+a2+…+an)(b1+b2+…+bn),求数列{cn}的前n项和Sn.解:(1)设{an}的公差为d,{bn}的公比为q.由2b2=a1+a2,2a2=b2+b3,得q=1,d=0或q=2,d=2.又a2≠b2,∴q=2,d=2,∴an=2n-1,bn=2n-1.(2)设{an}的前n项和为Mn,{bn}的前n项和为θn,由(1)得Mn=a1+a2+…+an=·n=n2,θn=b1+b2+…+bn==2n-1,故cn=·n2·(2n-1)=n·2n-n,∴Sn=(1×21+2×22+…+n×2n)-(1+2+…+n).令Tn=1×21+2×22+3×23+…+n×2n,①2Tn=1×22+2×23+3×24+…+n×2n+1,②由②-①,得Tn=(n-1)·2n+1+2.∴Sn=(n-1)·2n+1-+2.导学号〚32470781〛
