已知关于x的二次方程anx^2-an+1x+1=0（n∈正整数）的两个根a,b满足6a-2ab+6b=3,且a1=7/6

1)a+b=a(n+1)/an,ab=1/an
6a-2ab+6b=[6a(n+1)-2]/an=3
a(n+1)=(3an+2)/6
2)a(n+1)-2/3=(3an+2)/6-2/3
a(n+1)-2/3=an/2-1/3
a(n+1)-2/3=1/2(an-2/3)
则数列{an-2/3}是等比数列
（3）an-2/3=(1/2)^(n-1)*(a1-2/3)=(1/2)^n
an=(1/2)^n+2/3
Sn=(1/2)+(1/2)^2+……+(1/2)^n+2/3*n
=(1/2)(1-2^-n)/(1-1/2)+2/3*n
=1-(1/2)^n+2n/3不好意思 题目打错了 请重新解一下谢谢nan=n(1/2)^n+2/3n分开来求先求n(1/2)^nSn=1*1/2+2*1/4+3*1/8+……+n(1/2)^n1/2Sn=1*1/4+2/1/8+……+(n-1)*(1/2)^n+n*(1/2)^(n+1)1/2Sn=1/2+1/4+1/8+……+(1/2)^n-n*(1/2)^(n+1)1/2Sn=(1/2)(1-2^-n)/(1-1/2)-n*(1/2)^(n+1)Sn=2-(1/2)^(n-1)-n*(1/2)^n再求2/3nTn=2/3+4/3+……+2n/3=（2/3+2n/3)*n/2=(1/3+n/3)*n=(n+n^2)/3所以总和=2-(1/2)^(n-1)-n*(1/2)^n+(n+n^2)/3