If $$N = \frac{1}{3} + \frac{1}{(3^2) }+ \frac{1}{(3^3)}$$, then N is between
-
A$$0$$ and $$\frac{1}{9}$$
-
B$$\frac{1}{9}$$ and $$\frac{1}{3}$$
-
C$$\frac{1}{3}$$ and $$\frac{8}{9}$$
-
D$$\frac{8}{9}$$ and $$\frac{4}{3}$$
-
E$$\frac{4}{3}$$ and $$2$$
显示答案
正确答案: C