If f(1) = 1 and $$f(n) = 2 \sum_{k-1}^{n-1} f(k),(n > 1)$$ then $$\sum_{k-1}^{m} f(k)$$ =

A

$$(3^{m}) - 1$$

B

$$(3^{m})$$

C

$$(3^{m} - 2)$$

D

$$(3^{m} - 1)$$