If $$2\tan^2\theta - 5\sec\theta = 1$$ has exactly 7 solutions in the interval $$\left[0, \frac{n\pi}{2}\right]$$, for the least value of $$n \in \mathbb{N}$$ then $$\sum_{k=1}^{n} \frac{k}{2^k}$$ is equal to :

A

$$\frac{1}{2^{15}}(2^{14} - 14)$$

B

$$\frac{1}{2^{14}}(2^{15} - 15)$$

C

$$1 - \frac{15}{2^{13}}$$

D

$$\frac{1}{2^{13}}(2^{14} - 15)$$