已知│a-c-2│+(3a-6b-7)²+│3b+3c-4│=0,

因为│a-c-2│+(3a-6b-7)²+│3b+3c-4│=0
而│a-c-2│≥0
(3a-6b-7)²≥0
│3b+3c-4│≥0
所以a-c-2=0
3a-6b-7=0
3b+3c-4=0
解得a=3
b=1/3
c=1
所以原式=[3*（3^n)^3] *(1/3)^(3*n+1)*(1)^(3n-1)-3^3 *(1/3*1)^2
=(3)^(3*n+1)*(1/3)^(3*n+1)* 1- 27*1/9
=1-3
= - 2