Animated Solution for Mathematics - Binomial Theorem: If the constant term in the expansion of $(1+2x-3x^3){\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9$ is $p$, then $108p$ is equal to

Welcome, my dear student. Today, we are not just solving a problem; we are embarking on a detective mission. We are given the expression $(1+2x-3x^3){\left(\frac{3}{2}{x}^2-\frac{1}{3x}\right)}^9$ and asked to find the constant term $p$.

This is a classic JEE Advanced challenge. It tests not just your algebraic stamina, but your ability to see the structure hidden beneath the symbols. Many students see this and immediately panic, thinking they need to expand the entire binomial. Let me tell you: never do that. We are going to use the precision of the general term formula to find exactly what we need.

The Anatomy of the Expression

Look at the expression again. We have a polynomial $(1+2x-3x^3)$ multiplying a binomial expansion. If we want the constant term of the entire product, we must consider how the terms from the polynomial interact with the terms from the binomial.

The constant term of the product is the sum of the products of terms that result in $x^0$. Specifically, we have three scenarios:

1. The constant $1$ from the polynomial must multiply with the $x^0$ term from the binomial. 2. The term $2x$ from the polynomial must multiply with the $x^{-1}$ term from the binomial (because $x\cdot x^{-1}=x^0$). 3. The term $-3x^3$ from the polynomial must multiply with the $x^{-3}$ term from the binomial (because $x^3\cdot x^{-3}=x^0$).

This is our roadmap. We don't need the whole expansion; we only need these three specific terms.

The General Term Formula

To find these specific terms, we invoke the general term formula for $(a+b{)}^n$, which is $T_{r+1}=\left(\genfrac{}{}{0ex}{}{n}{r}\right)a^{n-r}{b}^r$. Here, $n=9$, $a=\frac{3}{2}{x}^2$, and $b=-\frac{1}{3x}$. Let's substitute these into our formula:

Now, let's perform some algebraic surgery. We need to separate the constants from the variables. This is where most students make mistakes, so let's be deliberate. We distribute the powers:

Combining the powers of $x$ and the powers of $3$, we get:

Simplifying the exponents of $3$ and $x$ gives us our master equation:

This, my friend, is the key to the kingdom. The exponent of $x$ is $18-3r$. Now we can solve our three cases.

The Three Cases

Case 1: The $x^0$ term. We need $18-3r=0$, which implies $3r=18$, so $r=6$. Since $6$ is a valid integer, this term exists! Let's calculate the coefficient:

Case 2: The $x^{-1}$ term. We need $18-3r=-1$, which implies $3r=19$, so $r=19/3$. Since $r$ must be an integer, this term does not exist. It contributes nothing to our constant $p$.

Case 3: The $x^{-3}$ term. We need $18-3r=-3$, which implies $3r=21$, so $r=7$. This is a valid integer! Let's calculate the coefficient, remembering to include the $-3$ from the polynomial:

The Final Victory

Now, we simply sum our contributions to find $p$:

We have found $p=1/2$. The question asks for $108p$. Therefore, $108\cdot (1/2)=54$.

Look at that! We didn't need to expand anything. We used the logic of the binomial theorem to isolate exactly what we needed. This is the essence of JEE mathematics: not brute force, but elegant, calculated precision. You have done well.