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:

<p>$$\begin{aligned} & 2 \tan ^2 \theta-5 \sec \theta-1=0 \\ & \Rightarrow 2 \sec ^2 \theta-5 \sec \theta-3=0 \\ & \Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0 \\ & \Rightarrow \sec \theta=-\frac{1}{2}, 3 \\ & \Rightarrow \cos \theta=-2, \frac{1}{3} \\ & \Rightarrow \cos \theta=\frac{1}{3} \end{aligned}$$</p> <p>For 7 solutions $\mathrm{n}=13$</p> <p>$$\begin{aligned} & \text { So, } \sum_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S} \text { (say) } \\ & \mathrm{S}=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\ldots .+\frac{13}{2^{13}} \\ & \frac{1}{2} \mathrm{~S}=\frac{1}{2^2}+\frac{1}{2^3}+\ldots .+\frac{12}{2^{13}}+\frac{13}{2^{14}} \\ & \Rightarrow \frac{\mathrm{S}}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow \mathrm{S}=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}} \end{aligned}$$</p>