高三一轮数列 数列求和错位相减家庭作业高手进啊

an=2n-1,bn=2/3*[(1/3)^(n-1)]
∵cn=anbn
∴cn=2/3*(2n-1)(1/3)^(n-1)
∴Sn=2/3(1*1+3*1/3+5*(1/3)^2+……+(2n-3)*(1/3)^(n-2)+(2n-1)(1/3)^(n-1))
1/3Sn=2/3(1*1/3+3*(1/3)^2+5(1/3)^3+……+(2n-3)(1/3)^(n-1)+(2n-1)*(1/3)^n)
两式相减得到
2/3Sn=2/3(1+2*1/3+2*(1/3)^2+……+2*(1/3)^(n-1)-(2n-1)*(1/3)^n)
∴Sn=2(1+1/3+(1/3)^2+……+(1/3)^(n-1))-(2n-1)(1/3)^n-1
=3(1-(1/3)^n)-(2n-1)(1/3)^n-1
=2-(2n-1+3)(1/3)^n
=2-2(n+1)(1/3)^n

x^2-12x+27=0
(x-3)(x-9)=0
x=3或x=9
公差d大于0
所以a2=3,a5=9
a5=a2+3d
9=3+3d
d=2
a2=a1+d
3=a1+2
a1=1
an=a1+(n-1)d
=1+2(n-1)
=2n-1
t1=1-1/2b1
b1=1-1/...