已知正项数列{an}=1,前n项和Sn满足an=根号下Sn+根号下Sn-1(n大于等于2) 求证根号下Sn为等差数列

1.n≥2时,an=Sn-S(n-1)=√Sn+√S(n-1)[√Sn+√S(n-1)][√Sn-√S(n-1)]=√Sn+√S(n-1)[√Sn+√S(n-1)][√Sn-√S(n-1) -1]=0算术平方根恒非负,√Sn≥0,√S(n-1)≥0√Sn+√S(n-1)≥0,又√S1=√a1=√1=1>0,因此√Sn+√S(...