本题采用分类求和.
a1+a2+a3+a4+a5+a6+……+a(3n-2)+a(3n-1)+a3n
=[a1+a4+……+a(3n-2)]+[a2+a5+……+a(3n-1)]+(a3+a6+……+a3n)
=[1/5+1/5^4+……+1/5^(3n-2)]+[1/5²+1/5^5+……+1/5^(3n-1)]+[1/5³+1/5^6+……+1/5^3n]
=1/5·[1-1/5^(3n)]/(1-1/5³)+1/5²·[1-1/5^(3n)]/(1-1/5³)+1/5³·[1-1/5^(3n)]/(1-1/5³)
=｛[1-1/5^(3n)]/(1-1/5³)｝·(1/5+1/5²+1/5³)
=1/4·(1-1/125^n)