an>0 根号Sn=(a1+an)/2 求an

令n=2,则S2=4
又Sn递增，故an递增
所以a1=1 a2=3

又Sn=[(a1+an)/2]^2
Sn-1=[(a1+an-1)/2]^2 两式相减
所以an=(2a1+an+an-1)(an-an-1)/4
=(2+an+an-1)(an-an-1)/4
4an=an^2+2an-(an-1)^2-2(an-1)
即an^2-2an-(an-1)^2-2(an-1)=0
即(an+an-1)(an-an-1-2)=0
又an+an-1>0 所以an-an-1-2=0
所以an=an-1+2
所以Sn-a1==Sn-1+2n
所以an=2n-1