If 2 tan^2 θ-5 sec θ=1 has exactly 7 solutions in the interval [0, (n π)/2], for the least value of n ε N, then Σ_k =1^n k/(2^k) is equal to:

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