在数列an中,a1=1,当n大于2时,前n项和Sn满足Sn的平方=an(Sn-1/2) 求数列an的通项
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![在数列an中,a1=1,当n大于2时,前n项和Sn满足Sn的平方=an(Sn-1/2) 求数列an的通项](/uploads/image/z/2981713-49-3.jpg?t=%E5%9C%A8%E6%95%B0%E5%88%97an%E4%B8%AD%2Ca1%3D1%2C%E5%BD%93n%E5%A4%A7%E4%BA%8E2%E6%97%B6%2C%E5%89%8Dn%E9%A1%B9%E5%92%8CSn%E6%BB%A1%E8%B6%B3Sn%E7%9A%84%E5%B9%B3%E6%96%B9%3Dan%EF%BC%88Sn-1%2F2%EF%BC%89+%E6%B1%82%E6%95%B0%E5%88%97an%E7%9A%84%E9%80%9A%E9%A1%B9)
在数列an中,a1=1,当n大于2时,前n项和Sn满足Sn的平方=an(Sn-1/2) 求数列an的通项
在数列an中,a1=1,当n大于2时,前n项和Sn满足Sn的平方=an(Sn-1/2) 求数列an的通项
在数列an中,a1=1,当n大于2时,前n项和Sn满足Sn的平方=an(Sn-1/2) 求数列an的通项
Sn²=an(Sn-1/2)
an=Sn-Sn-1
Sn²=(Sn-Sn-1)(Sn-1/2)
=Sn²-SnSn-1-1/2*Sn+1/2*Sn-1
SnSn-1=-1/2(Sn-S-1)
2SnSn-1=Sn-1-Sn
两边同时除以SnSn-1
(1/Sn)-(1/Sn-1)=2
因为S1=1
S2=1/3
所以1/Sn是首项为1公差为2的等差数列
1/Sn=2n-1
Sn=1/2n-1
Sn-1=1/2n-3
∴an=Sn-Sn-1=(1/2n-1)-(1/2n-3)(可考虑通分)
s²n=(sn-s(n-1))(sn-1/2)
s²n=s²n-sn/2-s(n-1)*sn+s(n-1)/2
sn(1/2+s(n-1))=s(n-1)/2
sn=s(n-1)/(1+2s(n-1))
1/sn=(1+2s(n-1))/s(n-1)
1/sn=1/s(n-1)+2
1/sn=1/a1+2(n-1)
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s²n=(sn-s(n-1))(sn-1/2)
s²n=s²n-sn/2-s(n-1)*sn+s(n-1)/2
sn(1/2+s(n-1))=s(n-1)/2
sn=s(n-1)/(1+2s(n-1))
1/sn=(1+2s(n-1))/s(n-1)
1/sn=1/s(n-1)+2
1/sn=1/a1+2(n-1)
1/sn=2n-1
sn=1/(2n-1)
验证:a1=s1=1/(2×1)=1成立,
则an=sn-s(n-1)=-2/[(2n-1)(2n-3)];
即an=-2/[(2n-1)(2n-3)]
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扯淡~~高一学数列了吗??