已知数列{an}满足a1=根号2,前n项和Sn=1/2(an+2/an),求它的通项公式

Sn=1/2(an+2/an)=1/2[Sn-Sn-1+2/(Sn-Sn-1)] (n≥2 n∈N*)化简得Sn^2-Sn-1^2=2那么Sn-1^2-Sn-2^2=2……S2^2-S1^2=2利用累加法得Sn^2=S1^2+2(n-1)=2nSn=√2n an=Sn-Sn-1=√2n-√2n-2 (n≥2 n∈N*)将a1带入a1=√2 成立...