数列{an}前n项和为Sn,对任意n属于R,都有an>0且Sn=[(an-1)(an+2)]/2,求an通项公式
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数列{an}前n项和为Sn,对任意n属于R,都有an>0且Sn=[(an-1)(an+2)]/2,求an通项公式
![数列{an}前n项和为Sn,对任意n属于R,都有an>0且Sn=[(an-1)(an+2)]/2,求an通项公式](/uploads/image/z/18383975-71-5.jpg?t=%E6%95%B0%E5%88%97%EF%BD%9Ban%EF%BD%9D%E5%89%8Dn%E9%A1%B9%E5%92%8C%E4%B8%BASn%2C%E5%AF%B9%E4%BB%BB%E6%84%8Fn%E5%B1%9E%E4%BA%8ER%2C%E9%83%BD%E6%9C%89an%EF%BC%9E0%E4%B8%94Sn%3D%5B%28an-1%29%28an%2B2%29%5D%2F2%2C%E6%B1%82an%E9%80%9A%E9%A1%B9%E5%85%AC%E5%BC%8F)
1、当n=1时,有:
a1=[(a1-1)(a1+2)]/2
得:a1=2
2、当n≥2时,an=Sn-S(n-1),则:
an=[(an-1)(an+2)]/2-[a(n-1)-1]×[a(n-1)+2]/2
2an=[(an)²+an-2]-[a(n-1)²+a(n-1)-2]
[an²-a(n-1)²]-[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-1]=0
因an>0,则:an-a(n-1)=1=常数,则:
{an}是以a1=2为首项、以d=1为公差的等差数列,则an=n+1
a1=[(a1-1)(a1+2)]/2
得:a1=2
2、当n≥2时,an=Sn-S(n-1),则:
an=[(an-1)(an+2)]/2-[a(n-1)-1]×[a(n-1)+2]/2
2an=[(an)²+an-2]-[a(n-1)²+a(n-1)-2]
[an²-a(n-1)²]-[an+a(n-1)]=0
[an+a(n-1)][an-a(n-1)-1]=0
因an>0,则:an-a(n-1)=1=常数,则:
{an}是以a1=2为首项、以d=1为公差的等差数列,则an=n+1
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