Question

已知数列an中,an>0 且sn=1/2(an+n/an),求数列an的通项公式

Answer 1

当n=1时 s1=a1=1/2(a1+1/a1) 2a1^2=a1^2+1 a1=1 (an>0, 负值舍)
s(n-1)=1/2[a(n-1)+(n-1)/a(n-1)] sn-s(n-1)=an=1/2[an+n/an-a(n-1)-(n-1)/a(n-1)]
2an=an+n/an-a(n-1)-(n-1)/a(n-1) an=n/an-a(n-1)-(n-1)/a(n-1) an+a(n-1)=n/an-(n-1)/a(n-1)
an^2a(n-1)+ana(n-1)^2=na(n-1)-(n-1)an ( an^2-n)a(n-1)=-[a(n-1)^2-(n-1)]an
( an^2-n)/an=-[a(n-1)^2-(n-1)]/a(n-1) 所以(an^2-n)/an 是等比数列首项为0 所以 (an^2-n)/an =0
an^2-n=0 an=根号n