www.1862.net > 等差数列{An}是递增数列,前n项和为Sn,且A1,A2,...

等差数列{An}是递增数列,前n项和为Sn,且A1,A2,...

(1)由等差数列{an}是递增数列,可设{an}的公差为d(d>0),∵a1,a2,a5成等比数列,S5=a32,∴a22=a1a55a3=a32,解得a1=1d=2,∴an=2n-1.(2)假设存在正整数m,l,使数列am,am+l,am+2l为等比数列,则am+l2=amam+2l,而an=2n-1,∴[2(m+...

(I)设公差为d(d>0),则∵4S3=S6,a2+2是a1,a13的等比中项,∴4(3a1+3d)=6a1+15d(a1+d+2)2=a1(a1+12d)∴a1=1d=2或a1=?14d=?12∵d>0,∴a1=1d=2∴数列{an}的通项公式an=2n-1;(II)若存在m,k∈N*,使am+am+4=ak+2,则2m-1+2(m+4)-1=2...

1、设差为n, 比为q 则a2=2+n=2q a5=2+4n=2q^2 根据题意可得:n=2q-2 得 2+8q-8=2q^2 q=1或3 则n=0或4 根据题意n=4 An=2+4(n-1) 等差数列通项公式:an=a1+(n-1)*d,n为正整数 等比数列通项公式:an=a1q^n-1,公比q≠0,等比数列a1≠ 0。其中an中...

解:充分条件证明: n>=2 Sn=na1+n(n-1)d/2=(n^2)d+na1-nd/2 对Sn求导得Sn’=nd+a1-d/2=nd+a1+d-d-d/2=nd-3d/2+a2 因为a2>0, (n-3/2)>0,所以Sn>0 为单调递增数列。 必要条件证明: 因为Sn单调递增,所以Sn的导数大于0 即Sn’=nd+a1-d/2>0推出(n-1/...

∵a1=2,a1,a3,a7成等比数列∴a32=a1a7设等差数列的公差d,则(2+2d)2=2(2+6d),d>0∴d=1,an=n+1∵Sn=2n+1?2.∴b1=s1=2bn=sn-sn-1=2n+1-2-2n+2=2n(n≥2)当n=1时也适合∴bn=2n(2)∵cn=abn=2n+1∴Tn=(2+1)+(22+1)+…+(2n+1)=(2+22+23+…+2n...

(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+...

(1)等差数列{an}为递增数列,且a2,a5是方程x2-12x+27=0的两根,解方程x2-12x+27=0解得:x1=3,x2=9由题意得:a1=3,a2=9进而求得:an=2n-1.由Sn=3n?12当n=1时,b1=S1=1;当n≥2时,bn=Sn?Sn?1=3n?12?3n?1?12=3n-1.又因为b1=1适合公式,所...

(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)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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