1.已知关于x的二次方程anx²-a(n+1)x+1=0(n∈N)的两根x1,x2满足6(x1+x2)-2x1x2=3,且a1=1(1)试用an表示a(n+1)(2)求证:(an-2/3)是等比数列(3)求数列的通项公式an(4)求数列{an}的前n项和Sn注:anx²-a(n+1)x+1=0中anx^2的n为角标,下同 a(n+1)中n+1为角标,下同x1+x2中1与2都为角标,下同

问题描述:

1.已知关于x的二次方程anx²-a(n+1)x+1=0(n∈N)的两根x1,x2满足6(x1+x2)-2x1x2=3,且a1=1
(1)试用an表示a(n+1)
(2)求证:(an-2/3)是等比数列
(3)求数列的通项公式an
(4)求数列{an}的前n项和Sn
注:anx²-a(n+1)x+1=0中anx^2的n为角标,下同
a(n+1)中n+1为角标,下同
x1+x2中1与2都为角标,下同

6a(n+1)/an-2/an=3
a(n+1)=(1/2)*an+1/3
a(n+1)-2/3=(1/2)(an-2/3)
(an-2/3)是等比数列,首项是1/3公比是1/2
an-2/3=1/3*(1/2)^(n-1) ( a^(n-1)表示a的n-1次方)
an=1/3*(1/2)^(n-1)+2/3
Sn=1/3*(1-(1/2)^n)/(1-1/2)+2n/3
=(2/3)*(1-(1/2)^n)+2n/3
=2/3-(2/3)*(1/2)^n+2n/3