www.1862.net > 已知递增的等比数列{An}的前n项和Sn满足:S4=S1+28...

已知递增的等比数列{An}的前n项和Sn满足:S4=S1+28...

(I)∵S2=6,S4=301?q41?q2=1+q2∴a1(1?q2)1?q=6a1(1?q4)1?q=30两式相除可得,1?q41?q2=1+q2=5∵数列{an}递增,q>0∴q=2,a1=2∴an=2?2n?1=2n(II)∵bn=anlog12an=-n?2n∴Tn=?(1?2+2?22+…+n?2n)设Hn=1?2+2?22+…+n?2n2Hn=1?22+2?23+…+(n-1)?...

(1)设等比数列{an}的首项为a1,公比为q.依题意,有2(a3+2)=a2+a4,代入a2+a3+a4=28,得a3=8.∴a2+a4=20.∴a1q+a1q3=20 a3=a1q2=8 解之得q=2a1=2,或q=12a1=32又{an}单调递增,∴q=2,a1=2,∴an=2n,(2)bn=2n?log122n=-n?2n,∴-Sn=1...

(1)∵s3=12,即a1+a2+a3=12,∴3a2=12,a2=4.设数列{an}的公差为d(d>0),由题意得,a22=2a1?(a3+1),a22=2(a2-d)?(a2+d+1)得d=3或d=-4(舍),∴a1=a2-d=1,∴{an}的通项公式:an=3n-2.(2)bn=an3n=3n-23n=(3n-2)13n,∴Tn=1×13+4×132+7×133+...

设等比数列{a n }的公比为q,依题意有2(a 3 +2)=a 2 +a 4 ,(1)又a 2 +a 3 +a 4 =28,将(1)代入得a 3 =8.所以a 2 +a 4 =20.于是有 a 1 q+ a 1 q 3 =20 a 1 q 2 =8 解得 a 1 =2 q=2 或 a 1 =32 q= 1 2 又{a n }是递增的,故a 1 =2,q=2...

an=2的n次方(n=1、2、3.....);1/t1=1/2;1/t1+1/t2+1/t3+…+1/tn=4/lg2*(1-1/n)

A、∵{an}是等比数列,则由“a1<a2<a3”可得数列{an}是递增数列,故充分性成立.若数列{an}是递增数列,则一定有a1<a2<a3,故必要性成立.综上,“a1<a2<a3”是“数列{an}是递增数列”的充分必要条件,故A正确;B、若“a1<a3<a5”则q2>1,q>1或...

(Ⅰ)设{an}的公差为d,{bn}的公比为q,则a2b2=(3+d)q=12,①S3+b2=3a2+b2=3(3+d)+q=9+3d+q=20,即3d+q=11,变形可得q=11-3d,②代入①可得:(3+d)(11-d)=33+2d-3d2=12,3d2-2d-21=0,(3d+7)(d-3)=0,又由{an}是单调递增的等差数列,有...

(Ⅰ)∵递增等比数列{an}的前n项和为Sn,且a2=3,S3=13,∴a2=3s3=a1+a2+a3=13,解得q=3或q=13,∵数列{an}为递增等比数列,所以q=3,a1=1.∴{an}是首项为1,公比为3的等比数列.∴an=3n-1.…(3分)∵点P(bn,bn+1)在直线x-y+2=0上,∴bn+1-bn=2...

(Ⅰ)设公差为d,公比为q,则a2b2=(3+d)q=12①S3+b2=3a2+b2=3(3+d)+q=20②联立①②可得,(3d+7)(d-3)=0∵{an}是单调递增的等差数列,d>0.则d=3,q=2,∴an=3+(n-1)×3=3n,bn=2n-1…(6分)(Ⅱ)bn=2n-1,cn=n?2n-1,∴Tn=c1+c2+…+cnTn=1?20+...

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