Animated Solution for Mathematics - Binomial Theorem: The Coefficient of $x^{-6}$ in the expansion of $(\frac{4x}{5}+\frac{5}{2x^2}{)}^9$ is ______.

Unlocking the Coefficient

Welcome, fellow traveler on the JEE Advanced path. Today, we are not just solving a math problem; we are peeling back the layers of a binomial expansion to find a hidden treasure.

We are looking for the coefficient of $x^{-6}$ in the expansion of $(\frac{4x}{5}+\frac{5}{2x^2}{)}^9$. It might look like a daunting mountain of algebra, but with the right tools, it is merely a scenic walk.

The Magic Key

When faced with a binomial raised to a power like $9$, the amateur tries to expand everything. The master, however, uses the General Term Formula. Think of this formula as a surgical tool that allows us to extract any specific term from the expansion without touching the others.

The general term $T_{r+1}$ is defined as:

Here, our $n$ is $9$, our $a$ is $\frac{4x}{5}$, and our $b$ is $\frac{5}{2x^2}$. By substituting these, we are essentially setting up our chessboard. We aren't calculating yet; we are simply placing our pieces in their correct positions.

The Art of Separation

Now, we must perform a crucial maneuver: separating the constants from the variables. This is where most students stumble, but you won't. We break down the expression into its constituent parts:

By grouping the constants $\left(\genfrac{}{}{0ex}{}{9}{r}\right){\left(\frac{4}{5}\right)}^{9-r}{\left(\frac{5}{2}\right)}^r$ at the front, we isolate the $x$ terms at the back. This separation is your best defense against algebraic chaos.

It allows us to focus purely on the power of $x$:

The Exponent Hunt

This is the moment of truth. The problem demands the coefficient of $x^{-6}$. Since our general term has $x^{9-3r}$, we simply equate the exponents:

Solving this linear equation is straightforward: $3r=15$, which gives us $r=5$. We have found our target! The $6^{th}$ term ($T_{5+1}$) is the one that holds the coefficient we seek.

The Elegant Cancellation

With $r=5$, we return to our constant block. We calculate $\left(\genfrac{}{}{0ex}{}{9}{5}\right){\left(\frac{4}{5}\right)}^4{\left(\frac{5}{2}\right)}^5$.

Using the symmetry property, $\left(\genfrac{}{}{0ex}{}{9}{5}\right)=\left(\genfrac{}{}{0ex}{}{9}{4}\right)=126$. Now, look at the powers of $2$ and $5$. By expressing $4$ as $2^2$, we get:

This simplifies beautifully: $2^{8-5}\cdot 5^{5-4}=2^3\cdot 5^1=8\cdot 5=40$.

Finally, we multiply our binomial coefficient by this result: $126\cdot 40=5040$.

And there it is. The coefficient is $5040$. You didn't just solve a problem; you navigated the logic of binomials with precision. Keep this clarity, and no JEE problem will ever intimidate you again.