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为等差数列求an通项公式(2)记数列{1/an·an+1}的前n项和为Tn,若对任意的n属于N*,不等式4Tn
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![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为等差数列求an通项公式(2)记数列{1/an·an+1}的前n项和为Tn,若对任意的n属于N*,不等式4Tn](/uploads/image/z/2680285-13-5.jpg?t=%E5%B7%B2%E7%9F%A5%E6%AD%A3%E9%A1%B9%E6%95%B0%E5%88%97%7Ban%7D%3D1%2C%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3an%3D%E6%A0%B9%E5%8F%B7%E4%B8%8BSn%2B%E6%A0%B9%E5%8F%B7%E4%B8%8BSn-1%28n%E5%A4%A7%E4%BA%8E%E7%AD%89%E4%BA%8E2%29+%E6%B1%82%E8%AF%81%E6%A0%B9%E5%8F%B7%E4%B8%8BSn%E4%B8%BA%E7%AD%89%E5%B7%AE%E6%95%B0%E5%88%97%E6%B1%82an%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F%EF%BC%882%EF%BC%89%E8%AE%B0%E6%95%B0%E5%88%97%7B1%2Fan%C2%B7an%2B1%7D%E7%9A%84%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BATn%2C%E8%8B%A5%E5%AF%B9%E4%BB%BB%E6%84%8F%E7%9A%84n%E5%B1%9E%E4%BA%8EN%2A%2C%E4%B8%8D%E7%AD%89%E5%BC%8F4Tn)
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为等差数列求an通项公式(2)记数列{1/an·an+1}的前n项和为Tn,若对任意的n属于N*,不等式4Tn
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
求an通项公式
(2)记数列{1/an·an+1}的前n项和为Tn,若对任意的n属于N*,不等式4Tn
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为等差数列求an通项公式(2)记数列{1/an·an+1}的前n项和为Tn,若对任意的n属于N*,不等式4Tn
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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(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
(2)1/an·an+1=1/((2n-1)(2n+1))=1/2(1/(2n-1)-1/(2n+1))
Tn=1/2(1-1/3+1/3-1/5+......+1/(2n-1)-1/(2n+1))=1/2(1-1/(2n+1))<1/2
要使得不等式4Tn即a²-a-2>=0,(a-2)(a+1)>=0
a>=2或a<=-1
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