JEE Math Practice Question

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 $$n \in \mathbf{N}$$, then $$\sum\limits_{k=1}^n \frac{k}{2^k}$$ is equal to:

1. A.$$\frac{1}{2^{14}}\left(2^{15}-15\right)$$
2. B.$$1-\frac{15}{2^{13}}$$
3. C.$$\frac{1}{2^{15}}\left(2^{14}-14\right)$$
4. D.$$\frac{1}{2^{13}}\left(2^{14}-15\right)$$