If \( 2\tan^2\theta-5\sec\theta-1 \) has exactly \( 7 \) solutions in the interval \(\displaystyle \left[0,\frac{n\pi}{2}\right] \) for the least value of \( n\in \mathbb{N} \), then \(\displaystyle \sum_{k=1}^{n} \frac{k}{2^{k}} \) is equal to:
\(\displaystyle \frac{1}{2^{13}}\!\left(2^{14}-15\right) \)
\(\displaystyle \frac{1}{2^{15}}\!\left(2^{14}-14\right) \)
\( \displaystyle1-\frac{15}{2^{13}} \)
\( \displaystyle\frac{1}{2^{14}}\!\left(2^{15}-15\right) \)
