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
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/03 23:25:54
![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/11401304-32-4.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)
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
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
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=S1=(a1-1)*(a1+2)/2 解得a1=2 Sn=(an-1)*(an+2)/2 S(n-1)=[a(n-1)-1]*[a(n-1)+2] Sn-S(n-1)=an 化简得[an+a(n-1)]*[an-a(n-1)-1]=0 因为对于任意n属于R,an>0,即an+a(n-1)>0 所以an-a(n-1)-1=0, 所以数列为首项为2,公差为1的等差数列an=n+1...
全部展开
a1=S1=(a1-1)*(a1+2)/2 解得a1=2 Sn=(an-1)*(an+2)/2 S(n-1)=[a(n-1)-1]*[a(n-1)+2] Sn-S(n-1)=an 化简得[an+a(n-1)]*[an-a(n-1)-1]=0 因为对于任意n属于R,an>0,即an+a(n-1)>0 所以an-a(n-1)-1=0, 所以数列为首项为2,公差为1的等差数列an=n+1
收起