Animated Solution for Mathematics - Trigonometry: If $2{\tan }^2\theta -5\sec \theta =1$ has exactly 7 solutions in the interval $[0,\frac{n\pi }{2}]$, for the least value of $n\in \mathbb{N}$ then ${\sum }_{k=1}^n\frac{k}{2^k}$ is equal to :

Welcome, fellow traveler on the path to JEE excellence! Today, we are not just solving a problem; we are choreographing a dance between the periodic nature of trigonometry and the structured elegance of arithmetic-geometric progressions.

This problem is a classic because it tests your ability to bridge two seemingly unrelated worlds: the continuous world of waves and the discrete world of series.

We begin with the equation $2{\tan }^2\theta -5\sec \theta =1$. At first glance, it looks messy because we have both tangent and secant functions.

Remember your fundamental identity: ${\tan }^2\theta ={\sec }^2\theta -1$. By substituting this, we transform the entire equation into a quadratic in terms of $\sec \theta$:

Expanding this, we get $2{\sec }^2\theta -5\sec \theta -3=0$. Now, factorizing this is like finding the rhythm in a piece of music.

We split the middle term to get $(2\sec \theta +1)(\sec \theta -3)=0$. This gives us two potential roots: $\sec \theta =-\frac{1}{2}$ and $\sec \theta =3$.

Stop! Here is where the trap lies. As we discussed, $\sec \theta$ cannot be $-\frac{1}{2}$ because the range of $\sec \theta$ is $(-\infty ,-1]\cup [1,\infty )$.

Thus, we discard the first root and embrace $\sec \theta =3$, which implies $\cos \theta =\frac{1}{3}$.

Imagine the graph of $y=\cos \theta$. We are looking for the intersection with the horizontal line $y=\frac{1}{3}$.

Let $\alpha ={\cos }^{-1}(\frac{1}{3})$. In the first cycle $[0,2\pi ]$, the solutions are $\alpha$ and $2\pi -\alpha$. As we move to the next cycle $[2\pi ,4\pi ]$, the solutions shift by $2\pi$, giving us $2\pi +\alpha$ and $4\pi -\alpha$.

If we continue this pattern, the 7th solution is ${\theta }_7=6\pi +\alpha$ and the 8th is ${\theta }_8=8\pi -\alpha$. We need exactly 7 solutions in the interval $[0,\frac{n\pi }{2}]$.

This means the 7th solution must be inside the interval, and the 8th must be outside:

Solving for $n$, we find $12+\frac{2\alpha }{\pi }\le n<16-\frac{2\alpha }{\pi }$. Since $0<\alpha <\frac{\pi }{2}$, the term $\frac{2\alpha }{\pi }$ is between 0 and 1. This forces $n$ to be 13.

The AGP Sum

Now that we have $n=13$, we must compute the sum $S={\sum }_{k=1}^{13}\frac{k}{2^k}$. This is an Arithmetic-Geometric Progression.

To solve it, we use the classic shift-and-subtract method. Write out the sum, multiply by the common ratio $\frac{1}{2}$, and shift the terms:

Subtracting these two equations yields:

The first part is a simple Geometric Progression with 13 terms. Using the sum formula $\frac{a(1-r^n)}{1-r}$, we get $1-\frac{1}{2^{13}}$.

Finally, solving for $S$ gives us the elegant result:

And there you have it! You have navigated the trigonometric constraints, solved the inequality, and mastered the series summation. Keep this confidence, and remember: every complex problem is just a series of simple steps waiting for you to connect them.