If ${\tan }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->5\sec <!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}=1$ has exactly 7 solutions in the interval $[0,\frac{n\mathrm{\pi <!--\; \pi \; -->}}{2}]$, for the least value of $n\in <!--\; \in \; -->N$, then $\underset{\mathrm{k}=1}{\overset{\mathrm{n}}{\sum <!--\; \sum \; -->}}\frac{\mathrm{k}}{2^{\mathrm{n}}}$ is equal to

The correct option is 91/2¹³.

Let us solve the trigonometric equation and find the value of $n$ step-by-step.

Step 1: Simplify the trigonometric equation
The given equation is:
${\tan }^2\theta -5\sec \theta =1$
Using the fundamental Pythagorean identity, we know that:
${\tan }^2\theta ={\sec }^2\theta -1$
Substitute this into the equation:
$({\sec }^2\theta -1)-5\sec \theta =1$
Rearranging the terms, we get a quadratic equation in terms of $\sec \theta$:
${\sec }^2\theta -5\sec \theta -2=0$

Step 2: Solve the quadratic equation
Let $x=\sec \theta$. The equation becomes:
$x^2-5x-2=0$
Using the quadratic formula, we find:
$x=\frac{5\pm \sqrt{{(-5)}^2-4\left(1\right)(-2)}}{2}=\frac{5\pm \sqrt{25+8}}{2}=\frac{5\pm \sqrt{33}}{2}$

Step 3: Analyze the feasibility of the roots
Since $\sqrt{33}\approx 5.74$, we have two possible values for $\sec \theta$:
1) $\sec \theta =\frac{5+\sqrt{33}}{2}$ (which satisfies $\left|\sec \theta \right|\ge 1$)
2) $\sec \theta =\frac{5-\sqrt{33}}{2}$ (which has no real solution since $\left|\sec \theta \right|<1$ is impossible for real angles)
Therefore, we only consider:
$\sec \theta =\frac{5+\sqrt{33}}{2}$

Step 4: Find the general behavior of the solutions
Since $\cos \theta >0$, the solutions for $\theta$ lie in the first and fourth quadrants.
Let the principal angle in the first quadrant be $\alpha \in (0,\frac{\pi }{2})$.
Listing the solutions in ascending order:
- 1st solution: ${\theta }_1=\alpha \in (0,\frac{\pi }{2})$
- 2nd solution: ${\theta }_2=2\pi -\alpha \in (\frac{3\pi }{2},2\pi )$
- 3rd solution: ${\theta }_3=2\pi +\alpha \in (2\pi ,\frac{5\pi }{2})$
- 4th solution: ${\theta }_4=4\pi -\alpha \in (\frac{7\pi }{2},4\pi )$
- 5th solution: ${\theta }_5=4\pi +\alpha \in (4\pi ,\frac{9\pi }{2})$
- 6th solution: ${\theta }_6=6\pi -\alpha \in (\frac{11\pi }{2},6\pi )$
- 7th solution: ${\theta }_7=6\pi +\alpha \in (6\pi ,\frac{13\pi }{2})$
- 8th solution: ${\theta }_8=8\pi -\alpha \in (\frac{15\pi }{2},8\pi )$

We require the interval $[0,\frac{n\pi }{2}]$ to contain exactly 7 solutions.
This means:
${\theta }_7\le \frac{n\pi }{2}<{\theta }_8$
Substituting the expressions for ${\theta }_7$ and ${\theta }_8$:
$6\pi +\alpha \le \frac{n\pi }{2}<8\pi -\alpha$
Since $\alpha \in (0,\frac{\pi }{2})$, we have:
$6\pi +\alpha >6\pi =\frac{12\pi }{2}$
Therefore, the least integer value of $n$ that satisfies $\frac{n\pi }{2}\ge 6\pi +\alpha$ is:
$n=13$
Checking this value: if $n=13$, then the upper limit of the interval is $\frac{13\pi }{2}=6.5\pi$.
Since ${\theta }_7=6\pi +\alpha <6.5\pi$ and ${\theta }_8=8\pi -\alpha >7.5\pi$, the interval $[0,\frac{13\pi }{2}]$ contains exactly 7 solutions. Thus, the least value of $n$ is indeed $13$.

Step 5: Compute the summation value
We need to calculate the sum:
$\sum _{k=1}^n\frac{k}{2^n}$
Substituting $n=13$ into the summation:
$\sum _{k=1}^{13}\frac{k}{2^{13}}=\frac{1}{2^{13}}\sum _{k=1}^{13}k$
Using the formula for the sum of the first $m$ natural numbers, $\sum _{k=1}^{m}k=\frac{m(m+1)}{2}$:
$\sum _{k=1}^{13}k=\frac{13\times 14}{2}=13\times 7=91$
Substituting this back into the expression, we get:
$\frac{91}{2^{13}}$