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已知正项数列an,其前n项和为Sn,若6Sn=an^2+3an+2,a1,a3,a11成等比
求an的通项公式
人气:335 ℃ 时间:2020-02-02 06:26:17
解答
如下:
6Sn=an^2+3an+2
6S(n-1)=[a(n-1)]^2+3a(n-1)+2
6Sn-6S(n-1)=6an=an^2+3an+2-{[a(n-1)]^2+3a(n-1)+2}
an-a(n-1)=3
{an}为等差数列,d=3
a1,a3,a11成等比数列
a3^2=a1×a11
A2=A1+d;A4=A1+3d;A9=A1+8d
a3=a1+2d=a1+6
a11=a1+10d=a1+60
(a1+6)^2=a1×(a1+60)
a1=3/4 d=3
所以,通项式为:
an=3/4 +(n-1)×3
an=3n-9/4额答案为3n-1你好,不好意思,我算a1那步60那里应该是30.(马虎了)6Sn=an^2+3an+26S(n-1)=[a(n-1)]^2+3a(n-1)+26Sn-6S(n-1)=6an=an^2+3an+2-{[a(n-1)]^2+3a(n-1)+2}an-a(n-1)=3{an}为等差数列,d=3a1,a3,a11成等比数列a3^2=a1×a11a3=a1+2d=a1+6a11=a1+10d=a1+30(a1+6)^2=a1×(a1+30)a1=2 d=3所以,通项式为:an=2 +(n-1)×3an=3n-1如上,这次对了。嘎嘎嘎,不好意思O(∩_∩)O~
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