Question

已知数列an中，an大于0且Sn=1/2（an+n/an）求数列an的通项公式，急求！

Answer 1

您好，首先把n=1带入原式,
得a1=1,.Sn=1/2+n/2an,S(n+1)=1/2+(n+1)/2a(n+1).
S(n+1)-Sn=(n+1)/2a(n+1)-n/2an.
2a(n+1)=a(n+1)-na(n+1).
(n+2)a(n+1)=(n+1)an
[a(n+1)]/an=(n+1)/(n+2)
[a(n+1)]/an*an/[a(n-1)]*...*a2/a1=[a(n+1)]/a1=(n+1)/(n+2)*n/(n+1)*...*2/3
a(n+1)=(n+1)/3
所以an=n/3

Answer 2

∵Sn=1/2(Sn-S(n-1)+n/(Sn-S(n-1))
∴Sn^2-S(n-1)^2=n
数列{Sn^2-S(n-1)^2}是自然数数列
其前n项和=(S2^2-S1^2)+(S3^2-S2^2)+...+(Sn^2-S(n-1))=Sn^2-S1^2=2+3+...+n=n(n+1)/2-1
∴Sn^2=n(n+1)/2
Sn=√(n(n+1)/2) (an>0=>Sn>0)
an=Sn-S(n-1)=√(n(n+1)/2)-√(n(n-1)/2)

Answer 3