已知数列{an}的各项均为正数,前n项和为Sn,且Sn=an(an+1)/2,设bn=1/2Sn,Tn=b1+b2+…+
已知数列{an}的各项均为正数,前n项和为Sn,且Sn=an(an+1)/2,设bn=1/2Sn,Tn=b1+b2+…+bn,求Tn
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sn=an(an+1)/2---------------1
s(n-1)=a(n-1) (a(n-1)+1)/2-----------------2
1减2得
an = an(an+1)/2-a(n-1) (a(n-1)+1)/2
an^2-an-a^2(n-1) -a(n-1) =0
(an-a(n-1))(an+a(n-1))-(an+a(n-1))=0
(an-a(n-1)-1)(an+a(n-1))=0
∵an的各项均为正数
∴an-a(n-1)-1=0
即an-a(n-1)=1
∴是等差数列
a1=a1(a1+1)/2 a1=1 由上得到an=n
bn=1/2sn=1/an(an+1)=1/an-1/（an+1)=1-1/2
∴Tn=1-1/2+1/2-1/3+1/3-1/4+...+1/an-1/an+1
=1-1/an+1
=an/an+1
=n/n+1
s(n-1)=a(n-1) (a(n-1)+1)/2-----------------2
1减2得
an = an(an+1)/2-a(n-1) (a(n-1)+1)/2
an^2-an-a^2(n-1) -a(n-1) =0
(an-a(n-1))(an+a(n-1))-(an+a(n-1))=0
(an-a(n-1)-1)(an+a(n-1))=0
∵an的各项均为正数
∴an-a(n-1)-1=0
即an-a(n-1)=1
∴是等差数列
a1=a1(a1+1)/2 a1=1 由上得到an=n
bn=1/2sn=1/an(an+1)=1/an-1/（an+1)=1-1/2
∴Tn=1-1/2+1/2-1/3+1/3-1/4+...+1/an-1/an+1
=1-1/an+1
=an/an+1
=n/n+1
1年前
8
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