设有数列an,a1=5/6,若以a1,a2,...an为系数的二次方程:an-1X^2-anx+1=0,都有根A,B,且3A-AB+3B=1
来源:百度知道 编辑:UC知道 时间:2024/06/01 04:53:52
求证{an-1/2}为等比数列
因为an-1X^2-anx+1=0,所以A+B=an/an-1 AB=1/an-1 又因为3A-AB+3B=1,
所以n大于等于2时
(3an -1)/an-1=1 即3an-1=an-1所以
(an-1/2)/(an-1 -1/2)=1/3
所以{an-1/2}为等比数列
A+B = An/An-1
AB = 1/An-1
3(An/An-1) - 1/An-1 =1
=>
3An - 1 =An-1
3(An - 1/2) = An-1 - 1/2
故q = 1/3
已知数列{an}满足 a1=1/2 , a1+a2+...+an=n^2an
已知数列an+1=an/(2an*an+1) a1=1 求an的通项公式
已知数列{an},a1=-7,,an+1=an+2,,求a1+a2+......a17=
已知数列An中,a1=1,an+1=2(a1+a2+...+an)
已知数列an满足a1=0.5,an=(an-1)+1/(n^2-1),则数列an的通项公式为?
数列{an}满足a1=1 a n+1=1/2an+1/2^n,求通项 an
已知:数列{an},满足a1=2,[a(n+1)]/an=n/(n+1),则通项an=
数学题 已知数列{an}满足a1=5,a2=5,an+1=an+6an-1若数列{an+1+tan}是等比数列.求数列{an}的通项公式
数列{An}中,A1=1,An=0.5An-1 -0.5,,则An=_________.
数列{an}满足a1=1,an+1=2an+n,求an.