考点规范练31 数列求和一、非标准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}:,…,+…+,…,若bn=,那么数列{bn}的前n项和Sn等于( )A.B.C.D.3.(2022山东济南模拟)在数列{an}中,an+1+(-1)nan=2n-1,则数列{an}的前12项和等于( )A.76B.78C.80D.824.已知等比数列{an}中,a1=3,a4=81,若数列{bn}满足bn=log3an,则数列的前n项和Sn= . 5.已知数列{an}的首项a1=3,通项an=2np+nq(n∈N+,p,q为常数),且a1,a4,a5成等差数列.求:(1)p,q的值;(2)数列{an}的前n项和Sn的公式.6.(2022广东惠州调研)已知向量p=(an,2n),向量q=(2n+1,-an+1),n∈N+,向量p与q垂直,且a1=1.(1)求数列{an}的通项公式;(2)若数列{bn}满足bn=log2an+1,求数列{an·bn}的前n项和Sn.7.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.-5-\n8.(2022山东,文19)在等差数列{an}中,已知公差d=2,a2是a1与a4的等比中项.(1)求数列{an}的通项公式;(2)设bn=,记Tn=-b1+b2-b3+b4-…+(-1)nbn,求Tn.9.已知等差数列{an}满足a2=0,a6+a8=-10.(1)求数列{an}的通项公式;(2)求数列的前n项和.10.设数列{an}满足a1=2,an+1-an=3·22n-1.(1)求数列{an}的通项公式;(2)令bn=nan,求数列{bn}的前n项和Sn.-5-\n11.已知数列{an}的前n项和Sn=-n2+kn(其中k∈N+),且Sn的最大值为8.(1)确定常数k,并求an;(2)求数列的前n项和Tn.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.##一、非标准1.A 解析:该数列的通项公式为an=(2n-1)+,则Sn=+=n2+1-.2.B 解析:易得an=,∴bn==4.∴Sn=4=4.3.B 解析:由已知an+1+(-1)nan=2n-1,得an+2+(-1)n+1an+1=2n+1,得an+2+an=(-1)n(2n-1)+(2n+1),取n=1,5,9及n=2,6,10,结果相加可得S12=a1+a2+a3+a4+…+a11+a12=78.故选B.4. 解析:设等比数列{an}的公比为q,则=q3=27,解得q=3.所以an=a1qn-1=3×3n-1=3n,故bn=log3an=n,所以,则数列的前n项和为1-+…+=1-.5.解:(1)由a1=3,得2p+q=3.又因为a4=24p+4q,a5=25p+5q,且a1+a5=2a4,得3+25p+5q=25p+8q,解得p=1,q=1.(2)由(1)知,an=2n+n,所以Sn=(2+22+…+2n)+(1+2+…+n)=2n+1-2+.6.解:(1)∵向量p与q垂直,∴2n+1an-2nan+1=0,即2nan+1=2n+1an.-5-\n∴=2.∴{an}是以1为首项,2为公比的等比数列.∴an=2n-1.(2)∵bn=log2an+1=n-1+1=n,∴an·bn=n·2n-1.∴Sn=1+2×2+3×22+4×23+…+n·2n-1.①∴2Sn=1×2+2×22+3×23+…+(n-1)·2n-1+n·2n.②①-②得,-Sn=1+2+22+23+24+…+2n-1-n·2n=-n·2n=(1-n)·2n-1,∴Sn=1+(n-1)·2n.7.解:(1)∵=an,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=+…+=.8.解:(1)由题意知(a1+d)2=a1(a1+3d),即(a1+2)2=a1(a1+6),解得a1=2,所以数列{an}的通项公式为an=2n.(2)由题意知bn==n(n+1),所以Tn=-1×2+2×3-3×4+…+(-1)nn·(n+1).因为bn+1-bn=2(n+1),可得当项数为偶数时,Tn=(-b1+b2)+(-b3+b4)+…+(-bn-1+bn)=4+8+12+…+2n=,当项数为奇数时,Tn=Tn-1+(-bn)=-n(n+1)=-.所以Tn=9.解:(1)设等差数列{an}的公差为d.由已知条件可得解得故数列{an}的通项公式为an=2-n.(2)设数列的前n项和为Sn,即Sn=a1++…+,故S1=1,+…+.所以,当n>1时,=a1++…+-5-\n=1-=1-.所以Sn=.综上,数列的前n项和Sn=.10.解:(1)由已知,当n≥1时,an+1=+a1=3(22n-1+22n-3+…+2)+2=22(n+1)-1.而a1=2,所以数列{an}的通项公式为an=22n-1.(2)由bn=nan=n·22n-1知Sn=1·2+2·23+3·25+…+n·22n-1.①从而22·Sn=1·23+2·25+3·27+…+n·22n+1.②①-②,得(1-22)Sn=2+23+25+…+22n-1-n·22n+1,即Sn=.11.解:(1)当n=k∈N+时,Sn=-n2+kn取得最大值,即8=Sk=-k2+k2=k2,故k2=16,即k=4.当n=1时,a1=S1=-+4=;当n≥2时,an=Sn-Sn-1=-n.当n=1时,上式也成立.综上,an=-n.(2)因为,所以Tn=1++…+,①所以2Tn=2+2++…+,②②-①得,2Tn-Tn=2+1++…+=4-=4-,故Tn=4-.12.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(n∈N+).∴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-.-5-
