Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^5$ in the expansion of $(2x^3-\frac{1}{3x^2}{)}^5$ is

Imagine you are standing before the expression $(2x^3-\frac{1}{3x^2}{)}^5$. If you were to expand this manually, you would be looking at six distinct terms, each with its own coefficients, powers, and potential for arithmetic errors.

It is a path filled with pitfalls. But as a student of mathematics, you know that algebra prefers elegance over brute force. We do not need to expand the entire expression to find the coefficient of $x^5$; we only need to find the specific term that contains it.

Your Mathematical Swiss Army Knife

We start with the general term formula:

This formula is the heartbeat of binomial expansions. It allows us to zoom in on any specific part of the expansion without looking at the rest.

In our case, $n=5$, $a=2x^3$, and $b=-\frac{1}{3x^2}$. Notice how we carefully include the negative sign with the $b$ term, as dropping it will cause the entire calculation to collapse.

Substituting these into our formula, we get:

Now, we must isolate the variable $x$. We separate the constants from the variables by pulling out the coefficients: $\left(\genfrac{}{}{0ex}{}{5}{r}\right)$, $2^{5-r}$, and $(-\frac{1}{3}{)}^r$.

What remains are the powers of $x$: $(x^3{)}^{5-r}$ and $(\frac{1}{x^2}{)}^r$. Using the laws of exponents, $(x^3{)}^{5-r}$ becomes $x^{15-3r}$, and $(\frac{1}{x^2}{)}^r$ becomes $x^{-2r}$.

When we multiply these, we add the exponents: $15-3r-2r=15-5r$. We have successfully condensed the entire variable structure into a single term: $x^{15-5r}$.

Our goal is to find the coefficient of $x^5$. This means the net exponent of $x$ must be exactly $5$.

We set up the equation:

Solving for $r$ is straightforward: $5r=10$, which gives us $r=2$. We have identified exactly which term in the expansion holds the key to our answer: the third term ($T_{2+1}=T_3$).

Now that we know $r=2$, we return to our constant coefficients. We substitute $r=2$ into our expression:

Let us break this down with precision: 1. $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=\frac{5\times 4}{2\times 1}=10$ 2. $2^{5-2}=2^3=8$ 3. $(-\frac{1}{3}{)}^2=\frac{1}{9}$

Multiplying these together, we get:

Look at what we have achieved. We did not expand the binomial, nor did we get lost in a sea of terms. We used the structure of the Binomial Theorem to surgically extract the answer.

This is the mindset of a JEE Advanced topper: identifying the most efficient path, respecting the algebra, and executing with precision. The final coefficient is $\frac{80}{9}$.