www.rtmj.net > 数列{An}和{Bn}的各项均为正数,且对于任意n∈N*,An+12=AnAn+2+(A2013%A...

数列{An}和{Bn}的各项均为正数,且对于任意n∈N*,An+12=AnAn+2+(A2013%A...

(1)由于对于任意n∈N*,an+12=anan+2+(a2013-a2012)2,令n=2011,则a20122=a2011a2013+a20132+a20122-2a2013a2012,化简得,a2011+a2013a2012=2,令n=2012,则a20132=a2012a2014+a20132+a20122-2a2013a2012,化

(1)证明:an+an+1=2bn,①bnbn+1=an+12,②②式两边开方得:an+1= bnbn+1 = bn ? bn+1 ,③①式两边平方,展开,然后将③代入,得:bnbn-1+bnbn+1+2 bn?1?bn?bn?bn+1 =4bn?bn,④整理,得 bn?1 + bn+1 =2 bn ,∴数列{本回答由提问者推荐答案纠错|评论素颜NKOL采纳率:72%擅长:暂未定制为您推荐:其他类似问题

(1)2Sn=an+an①2S(n-1)=a(n-1)+a(n-1)②②-①2Sn-2S(n-1)=an+an-a(n-1)-a(n-1)2an=an+an-a(n-1)-a(n-1)an+a(n-1)=an-a(n-1)∵数列{an}的各项均为正数

an+an^2=2Snan-1+an-1^2=2Sn-1 相减an+an^2-an-1-an-1^2=2an(an+an-1)(an-an-1)=an+an-1 {an}的各项均为正数an-an-1=1n=1 a1+a1^2=2a1 a1=0(舍)或a1=1{an}是以a1=1为首项,公差d=1的等差数列a

(1)令n=1,则a13=S12+2S1,即a13=a12+2a1,∴a1 =2或a1=-1或a1=0,又∵数列{an}的各项都是正数,∴a1=2,令n=2,则a13+a23=S22+2S2,即a13+a23=(a1+a2)2+2(a

由已知得2Sn=an+an2,①当n≥2时,2Sn-1=an-1+a2n-1,②①-②,得2an=an-an-1+a2n-a2n-1,即(an+an-1)(an-an-1-1)=0,∵数列{an}的各项均为正数,∴an-an-1=1,又n=1时,2a1=a1+a21,解得a1=1,∴{an}是首项为1,公差为1的等差数列,∴an=n.∴bn=1(2n+1)(2n+3)=12(12n+1-12n+3),∴Tn=b1+b2+…+bn=12(13-15+15-17+…+12n+1-12n+3).=12(13-12n+3)=n3(2n+3)=n6n+9.故选C.

a1^3+a2^3+a3^3+.+an^3=Sn^2 ①a1^3+a2^3+a3^3+.+an^3+a(n+1)^3=S(n+1)^2 ②②-- ① a(n+1)^3==S(n+1)^2-Sn^2a(n+1)^3==[S(n+1)-Sn][S(n+1)+Sn]a(n+1)^2==S(n+1)+Sn a(n+1)^2==a(n+1)+2Sn ③2S(n-1)==an

∵2Sn=3an-3∴当n≥2时 an=Sn-S(n-1)∴2an=3an-3a(n-1)整理得 an=3a(n-1)∵a1=S1 ∴a1=3∴数列{an}是公比为3,首项为3的等比数列 an=3*3^(n-1)=3^n∵bn=1/log3 an*log3 ^a(n+1)∴bn=1/n*(n+1)=1/n-1/(n+1)Tn=b1+b2+b3+.bn=1-1/2+1/2-1/3+1/3-1/4.1/(n-1)-1/n+1/n-1/(n+1)=1-1/(n+1)

(1)依题意an+an+1=2b2n(1)a2n+1=b2nb2n+1(2)(2分)∴bn-1+bn+1=2bn(n>1)∴{bn}为等差数列 (6分)(2)由a1=1,b1=2,求得bn=22(n+1)(8分)∴an=12n(n+1)∴Sn=1a1+1a2+…+1an=2(112+1213+…+1n1n+1)=2nn+1(12分)

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