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已知公差不为0的等差数列{An}满足A1,A3,A4成等比...

(Ⅰ)设公差为d,∵公差不为零的等差数列{an},满足a3=5且a1,a2,a4成等比数列.∴a1+2d=5(a1+d)2=a1(a1+3d)a1≠0,解得:a1=53,d=53,得an=53n(n∈N*)(Ⅱ)由题意an=53n,∴bn=153n?53(n+1)=925(1n?1n+1),∴Tn=925[(1?12)+(12?13)+…+(1n?1...

解答:(1)解:设公差为d≠0,∵a3=6,且a1,a2,a4成等比数列,∴a1+2d=6,且(a1+d)2=a1?(a1+3d),解得a1=2,d=2.∴数列{an}的通项公式为an=2+(n-1)×2=2n;∵bn+1=2bn+1,∴bn+1+1=2(bn+1),∵b1=3,∴数列{bn+1}是以4为首项,2为公比的等比...

(本小题满分12分)解:(Ⅰ)设数列{an}的公差为d(d≠0),∵S3=a4+2,∴3a1+3×2×d2=a1+3d+2.①…(3分)又∵a1,a2-1,a3-1成等比数列,∴a1(a1+2d?1)=(a1+d?1)2.②…(5分)由①②解得a1=1,d=2.…(6分)∴an=a1+(n-1)d=2n-1.…(7分)(Ⅱ)∵1anan...

∵等差数列{an}的公差和首项都不等于0,且a2,a4,a8成等比数列,∴a42=a2a8,∴(a1+3d)2=(a1+d)(a1+7d),∴d2=a1d,∵d≠0,∴d=a1,∴a1+a5+a9a2+a3=15a15a1=3.故选:B.

数学知识都还给老师了哈哈,回答你第一问。 a1+a2=2a1+d=1 a2,a3,a5成等比数劫,a5/a3=a3/a2,即(a1+4d)/(a1+2d)=(a1+2d)/(a1+d),化简这个等式,得到a1d=0,因为d不等于0,所以a1=0。 2a1+d=1,得知d=1 所以数列{an}的通项公式为:an=a1+(n-1...

∵数列{an}是一个公差不为0等差数列,且a2=2,并且a3,a6,a12成等比数列,∴a62=a3?a12,∴(2+4d)2=(2+d)(2+10d),∵d≠0,∴d=1.∴an=2+(n-2)=n,∴1anan+1=1n-1n+1,∴1a1a2+1a2a3+1a3a4+…+1anan+1=1-12+12-13+…+1n-1n+1=1-1n+1=nn+1,故答案...

解: 设公差为d,则d≠0。 a4是a3、a7的等比中项,则 a4²=a3·a7 (a3+d)²=a3·(a3+4d) 整理,得2a3d-d²=0 d(2a3-d)=0 d=0(舍去)或2a3-d=0 a3=d/2 a1=a3-2d=d/2 -2d=(-3/2)d S8=8a1+8×7d/2=8·(-3/2)d+28d=16d=32 d=2 a1=(-3/2)d=(-3/...

解: 设公比为q,数列是正项数列,则数列各项均为正,且公比q>0 由等比中项性质得a3²=a1·a5 a4²=9a1a5=9a3² a4>0,a3>0,因此a4=3a3 q=a4/a3=3 a1、2a2、a3+6成等差数列,则2·(2a2)=a1+a3+6 4a1q=a1+a1q²+6 q=3代入,整理,...

a3、a5、a8成等比数列,则a5²=a3·a8 (a1+4d)²=(a1+2d)(a1+7d) 2d²-da1=0 a1=2代入,得d²-d=0 d(d-1)=0 d=0(舍去)或d=1 an=a1+(n-1)d=2+1·(n-1)=n+1 数列{an}的通项公式为an=n+1

(1)设等差数列{an}的公差为d,则5a1+5×42×d=20(a1+2d)2=a1?(a1+6d),解得d=1a1=2,∴an=2+n-1=n+1.(2)由(1)得,1anan+1=1(n+1)(n+2)=1n+1?1n+2,则Tn=(12?13)+(13?14)+…+(1n+1?1n+2)=12?1n+2=n2(n+2),∴Tn≤λan+1对一切n∈N*恒成...

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