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)An=3n-1 Bn=2^n (2)Tn=(3n-1)*2+(3n-4)*2^2+(3n-7)*2^3+......+8*2^(n-2)+5*2^(n-1)+2*2^n ① 2Tn=(3n-1)*2^2+(3n-4)*2^3+(3n-7)*2^4+......+8*2^(n-1)+5*2^n+2*2^(n+1)② ①-②,得Tn=-(3n-1)*2 + 3[2^2 + 2^3 + ... + 2^n] + 2^(n+2) =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...

(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+...

若an为递增数列 则有an

(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+...

(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...

(1)①∵等差数列{an}为递增数列,且a2,a5是方程x2-12x+27=0的两根,∴a2+a5=12,a2a5=27,∵d>0,∴a2=3,a5=9,∴d=a5?a23=2,a1=1,∴an=2n-1(n∈N*)②∵Tn=1-12bn,∴令n=1,得b1=23,当n≥2时,Tn=1-12bn,Tn-1=1-12bn-1,两式相减得,bn=12bn-1-1...

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