设数列{an}各项为正数,前n项和为Sn,且2*二倍根号下Sn=an+1,(n为一切正整数) (1)求数列{an}通项公式(2)记bn=1/(二倍根号下an+二倍根号下an+1),求数列{bn}的前n项和Tn.
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![设数列{an}各项为正数,前n项和为Sn,且2*二倍根号下Sn=an+1,(n为一切正整数) (1)求数列{an}通项公式(2)记bn=1/(二倍根号下an+二倍根号下an+1),求数列{bn}的前n项和Tn.](/uploads/image/z/1741124-20-4.jpg?t=%E8%AE%BE%E6%95%B0%E5%88%97%7Ban%7D%E5%90%84%E9%A1%B9%E4%B8%BA%E6%AD%A3%E6%95%B0%2C%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E4%B8%942%2A%E4%BA%8C%E5%80%8D%E6%A0%B9%E5%8F%B7%E4%B8%8BSn%3Dan%2B1%2C%EF%BC%88n%E4%B8%BA%E4%B8%80%E5%88%87%E6%AD%A3%E6%95%B4%E6%95%B0%EF%BC%89+%EF%BC%881%EF%BC%89%E6%B1%82%E6%95%B0%E5%88%97%7Ban%7D%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E8%AE%B0bn%3D1%2F%28%E4%BA%8C%E5%80%8D%E6%A0%B9%E5%8F%B7%E4%B8%8Ban%2B%E4%BA%8C%E5%80%8D%E6%A0%B9%E5%8F%B7%E4%B8%8Ban%2B1%29%2C%E6%B1%82%E6%95%B0%E5%88%97%7Bbn%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8CTn.)
设数列{an}各项为正数,前n项和为Sn,且2*二倍根号下Sn=an+1,(n为一切正整数) (1)求数列{an}通项公式(2)记bn=1/(二倍根号下an+二倍根号下an+1),求数列{bn}的前n项和Tn.
设数列{an}各项为正数,前n项和为Sn,且2*二倍根号下Sn=an+1,(n为一切正整数) (1)求数列{an}通项公式
(2)记bn=1/(二倍根号下an+二倍根号下an+1),求数列{bn}的前n项和Tn.
设数列{an}各项为正数,前n项和为Sn,且2*二倍根号下Sn=an+1,(n为一切正整数) (1)求数列{an}通项公式(2)记bn=1/(二倍根号下an+二倍根号下an+1),求数列{bn}的前n项和Tn.
题目应有笔误,应该是“设数列{an}各项为正数,前n项和为Sn,且二倍根号下Sn=an+1,(n为一切正整数) (1)求数列{an}通项公式(2)记bn=1/(二倍根号下an+二倍根号下an+1),求数列{bn}的前n项和Tn”吧?
2√S(n)=a(n)+1,得2√a(1)=a(1)+1,解得a(1)=1
并有4S(n)=[a(n)+1]^2,及4S(n+1)=[a(n+1)+1]^2.后者减前者,易得
4a(n+1)=a(n+1)^2+2a(n+1)-a(n)^2-2a(n)
也即a(n+1)^2-2a(n+1)-a(n)^2-2a(n)=0
[a(n+1)+a(n)]*[a(n+1)-a(n)]-2[a(n+1)+a(n)]=0
[a(n+1)+a(n)]*[a(n+1)-a(n)-2]=0
因数列a(n)各项为正,故a(n+1)+a(n)>0,则必有a(n+1)-a(n)-2=0.于是数列a(n)是首项为1,公差为2的等差数列.则a(n)=2n-1
则b(n)=1/[2√a(n)+2√a(n+1)]=1/[2√(2n-1)+2√(2n+1)]=[√(2n+1)-√(2n-1)]/4
则数列{bn}的前n项和Tn=[√(2n+1)-1]/4