已知正项数列{An}首项A1=1,前n项和Sn满足An=√Sn+√Sn-1(n≥2)求证{√Sn}为等差数列,并求An通

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已知正项数列{An}首项A1=1,前n项和Sn满足An=√Sn+√Sn-1(n≥2)求证{√Sn}为等差数列,并求An通项公式

数列为正项数列,则Sn>0
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,为定值.
√S1=√a1=√1=1
数列{√Sn}是以1为首项,1为公差的等差数列.
√Sn=1+(n-1)=n
Sn=n²
n≥2时,an=Sn-S(n-1)=n²-(n-1)²=2n-1
n=1时,a1=2×1-1=1,同样满足通项公式.
数列{an}的通项公式为an=2n-1.

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