设二次方程anx^2-a(n+1)x+1=0有两个根x1,x2,且满足6x1-2x1x2+6x2=3.且a1=1
问题描述:
设二次方程anx^2-a(n+1)x+1=0有两个根x1,x2,且满足6x1-2x1x2+6x2=3.且a1=1
设二次方程anx²-a(n+1)x+1=0(n=1,2,3…)有两根α和β,且满足6α-2αβ+6β=3。,a1=1
(1)试用an表示a(n+1);
(2)求证:{an-2/3}是等比数列;
(3)求数列{an}的通项公式。
答
6α-2αβ+6β=36(α+β)-2αβ=36a(n+1)/an -2/an=3a(n+1)=(1/2)an+(1/3)a(n+1)-(2/3)=(1/2)an+(1/3)-(2/3)=(1/2)[an-(2/3)]所以:{an-2/3}是公比为1/2的等比数列设bn=an-(2/3)则:b1=a1-(2/3)=1/3bn=b1*(1/2)^(n-1...