Sn=(an+1/an)
s1=a1=(a1+1/a1)/2
==>a1=1/a1
==>a1=1(由于an>0所以a1=-1不合题意)
s2=a1+a2=(a2+1/a2)/2
==>2a1+a2=1/a2
将a1=1代入得
2+a2=1/a2
==>a2^2+2a2-1=0
==(a2+1)^2-2=0
==>a2=√2-1
s3=a1+a2+a3=(a3+1/a3)/2
将a1=1 a2=√2-1 代入得
1+√2-1+a3=(a3+1/a3)/2
==>2√2+a3=1/a3
==>a3^2+2√2a3-1=0
==>(a3+√2)^2-2-1=0
==>a3=√3-√2
用归纳法
假设an=√n-√(n-1)
则sn=a1+a2+..+an=√n-1
当N=1时
Sn+1=an+Sn=(an+1/an+1)/2
==>an+√n-1=(an+1/an+1)/2
==>√n-√(n-1)+√n-1=(an+1/an+1)/2
==>an+1=√n+1-√n
故得证