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(n-1)不恒等于0,要等式成立,只有
√Sn-√S(n-1)-1=0
√Sn-√S(n-1)=1,为定值.
√S1=√a1=1,数列{√Sn}是以1为首项,1为公差的等差数列.
√Sn=1+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,同样满足通项公式
数列{an}的通项公式为an=2n-1
2.
1/[ana(n+1)]=1/[(2n-1)(2n+1)]=(1/2)[1/(2n-1) -1/(2n+1)]
4Tn=4[1/a1a2+1/a2a3+...+1/ana(n+1)]
=2[1/1-1/3+1/3-1/5+...+1/(2n-1)-1/(2n+1)]
=2[1-1/(2n+1)]
=4n/(2n+1)
=(4n+2-2)/(2n+1)
=2 -2/(2n+1)
随n增大,2n+1单调递增,2/(2n+1)单调递减,2- 2/(2n+1)单调递增,当n->+∞时,-2/(2n+1)->0
2- 2/(2n+1)->2
4Tn

(1)由an=√sn+√sn-1,an=sn-sn-1,可得
sn-sn-1=√sn+√sn-1
平方差公式sn-sn-1=(√sn+√sn-1)(√sn-√sn-1)=√sn+√sn-1
所以√sn-√sn-1=1
√sn为首项为1公差为1的等差数列，
所以√sn=n,sn=n^2,
an=sn-sn-1=n^2-(n-1)^2=2n-1
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