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.已知a1=7/6.(1)用an表示a(n+1)(2)求{an}的通项公式(3)求{an}的前n项之和sn
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.已知a1=7/6.
(1)用an表示a(n+1)
(2)求{an}的通项公式
(3)求{an}的前n项之和sn
答
1:由韦达定理得
x1+x2=a(n+1)/an
x1*x2=1/an
代入6x1-2x1x2+6x2=3并整理得
a(n+1) =an/2+1/3
2:a(n+1)=an/2+1/3,
[a(n+1)]-2/3=an/2+1/3 -2/3=an/2-1/3=(an-2/3)/2
{an - 2/3} 是以 1/2 为公比的等比数列
另有 a1 - 2/3 = 7/6 - 2/3 = 1/2
an - 2/3 = 1/2 * (1/2)^(n-1)=1/2^n
所以 an = 2/3 +1/2^n
3:Sn=a1+a2+...+an
=2n/3+(1/2+1/4+...+1/2^n)
=2n/3+[1/2-1/2^(n+1)]/(1-1/2)
=2n/3-1/2^n +1