数列{an}首项为2,且对任意n∈N*,都有1/a1a2+1/a2a3+...+1/anan+1=n/a1an+1,数列{an}前10项和为110(1)求证:数列{an}为等差数列;(2)设Cn=an•(1/2)^n,求数列{Cn}的前n项和Tn;(3)若存在n∈N*,使得an
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![数列{an}首项为2,且对任意n∈N*,都有1/a1a2+1/a2a3+...+1/anan+1=n/a1an+1,数列{an}前10项和为110(1)求证:数列{an}为等差数列;(2)设Cn=an•(1/2)^n,求数列{Cn}的前n项和Tn;(3)若存在n∈N*,使得an](/uploads/image/z/15230778-42-8.jpg?t=%E6%95%B0%E5%88%97%7Ban%7D%E9%A6%96%E9%A1%B9%E4%B8%BA2%2C%E4%B8%94%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E2%88%88N%2A%2C%E9%83%BD%E6%9C%891%2Fa1a2%2B1%2Fa2a3%2B...%2B1%2Fanan%2B1%3Dn%2Fa1an%2B1%2C%E6%95%B0%E5%88%97%7Ban%7D%E5%89%8D10%E9%A1%B9%E5%92%8C%E4%B8%BA110%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9A%E6%95%B0%E5%88%97%7Ban%7D%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%EF%BC%9B%EF%BC%882%EF%BC%89%E8%AE%BECn%3Dan%26%238226%3B%EF%BC%881%2F2%EF%BC%89%5En%2C%E6%B1%82%E6%95%B0%E5%88%97%7BCn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn%EF%BC%9B%EF%BC%883%EF%BC%89%E8%8B%A5%E5%AD%98%E5%9C%A8n%E2%88%88N%2A%2C%E4%BD%BF%E5%BE%97an)
数列{an}首项为2,且对任意n∈N*,都有1/a1a2+1/a2a3+...+1/anan+1=n/a1an+1,数列{an}前10项和为110(1)求证:数列{an}为等差数列;(2)设Cn=an•(1/2)^n,求数列{Cn}的前n项和Tn;(3)若存在n∈N*,使得an
数列{an}首项为2,且对任意n∈N*,都有1/a1a2+1/a2a3+...+1/anan+1=n/a1an+1,数列{an}前10项和为110
(1)求证:数列{an}为等差数列;
(2)设Cn=an•(1/2)^n,求数列{Cn}的前n项和Tn;
(3)若存在n∈N*,使得an≤(n+1)λ成立,求实数λ的最小值.
数列{an}首项为2,且对任意n∈N*,都有1/a1a2+1/a2a3+...+1/anan+1=n/a1an+1,数列{an}前10项和为110(1)求证:数列{an}为等差数列;(2)设Cn=an•(1/2)^n,求数列{Cn}的前n项和Tn;(3)若存在n∈N*,使得an
由题意得1/a1a2+1/a2a3…1/anan-1=(n-1)/a1an①
原式-①得
1/anan+1=n/a1an+1-(n-1)a1an
整理得2=nan-(n-1)an+1
两边同时除以n(n-1)得
2/n(n-1)=an/(n-1)-an+1/n
2/(n-1)-2/n=an/(n-1)-an+1/n
(An+1 -2)/n=(an -2)/(n-1)=…(a2-2)/1
An=(a2-2)(n-1)+2
A2是常数即an是等差数列
S10=(a1+a10)*10/2=110
A10=20
D=(a10-a1)/(10-1)=2
∴an=2n
Tnx2-tn的老路子 步骤不详写
tn=4-(2n+4)/2^n
入=2n/(n+1) n趋向无穷大入趋向2 即最小值为2