www.rtmj.net > 已知数列{An}是等差数列,其前n项和为Sn,若A1A2A3...

已知数列{An}是等差数列,其前n项和为Sn,若A1A2A3...

∵数列{an}是等差数列,∴S1=a1,S3=3a2,S5=5a3,又∵3S1S3+15S3S5+5S5S1=35∴1a1a2+1a2a3+1a1a3=35又∵a1a2a3=15∴a315+a115+a215=a25=35,∴a2=3故选C

设公差为x,则有 a1=1/2 s2=a3 2a1+x=a3 a3-a1-a1=2x-a1 a3-a1-a1=x 代入得x-a1=0 x=1/2 则公差为1/2 所以a2=a1+x=1 所以Sn=na1+n(n-1)x/2=n1/2+n(n-1)/4=n(n-1+2)/4=n(n+1)/4=(n^2+n)/4 那两楼对Sn的计算都是错误的,我验算了,我这答案应该是...

(1)∵{bn}是等比数列,首项为4,公比为2,∴bn=4?2n-1=2n+1,∵数列{an}是等差数列,且对任意的n∈N*,都有a1b1+a2b2+a3b3+…+anbn=n?2n+3,∴a1b1=24,∴a1=24b1=244=4,a1b1+a2b2=2?25,∴a2b2=2?25?24=48,∴a2=48b2=4823=6,∴d=a2-a1=6-4=2,...

设等差数列{an}的公差为d,则2d=(a2+a4)-(a1+a3)=10-4=6,解得d=3,∴a1+a3=a1+a1+2d=4,解得a1=2-d=-1,∴Sn=na1+n(n?1)2d=32n2?52n故答案为:32n2?52n

总数不变。 28400/71=400 n等于400

设{an}的公差为d,由题意得a1+a3+a5=a1+a1+2d+a1+4d=105,即a1+2d=35,①a2+a4+a6=a1+d+a1+3d+a1+5d=99,即a1+3d=33,②由①②联立得a1=39,d=-2,∴sn=39n+n(n?1)2×(-2)=-n2+40n=-(n-20)2+400,故当n=20时,Sn达到最大值400.故答案为:20

(1)∵等差数列{an}中,公差d>0,a2?a3=45,a1+a4=14,∴(a1+2d)(a1+2d)=45,a1+a1+3d=14∵d>0,∴a1=1,d=4,∴an=a1+(n-1)d=1+4(n-1)=4n-3;(2)Sn=n(1+4n?3)2=2n(n?12),bn=Snn+c=2n(n?12)n+c,令c=-12,即得bn=2n,{bn}为等差数列...

(1){an}为等差数列,所以a1+a4=a2+a3=14,又a2a3=45,所以a2,a3是方程x2-14x+45=0的两实根,公差d>0,∴a2<a3∴a2=5,a3=9∴a1+d=5a1+2d=9?a1=1d=4所以an=4n-3(2)由(1)知sn=2n2-n,所以bn=snn+c=2n2?nn+c∴b1=11+c,b2=62+c,b3=1...

(1)∵Sn=12n-n2.∴当n=1时,a1=S1=12-1=11,当n≥2时,an=Sn-Sn-1=(12n-n2)-12(n-1)+(n-1)2=13-2n.当n=1时,13-2×1=11=a1,∴an=13-2n.由an=13-2n≥0,得n≤132,∴当1≤n≤6时,an>0;当n≥7时,an<0.∴|a1|+|a2|+|a3|=S3=12×3-32=27.(2)...

(1)∵S3=18,a1+1,a2,a3成等比数列∴a22=(a1+1)a3∴3a1+3d=18(a1+d)2=(1+a1)(a1+2d)解可得,d=3或d=-2(舍去),a1=3∴an=3+3(n-1)=3n(2)∵bn=an3n+1=3n3n+1=n3n∴Tn=1?13+2?132+…+n?13n13Tn=1?132+2?133+…+n?13n+n3n+1两式相减可得,23Tn...

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