Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^2$ in the expansion of the product $(2-x^2)\cdot ((1+2x+3x^2{)}^6+(1-4x^2{)}^6)$ is :

Welcome, my dear student. Today, we are going to dismantle a problem that, at first glance, looks like a monster. You see an expression like $(2-x^2)\cdot ((1+2x+3x^2{)}^6+(1-4x^2{)}^6)$ and your instinct might be to panic.

You might think, 'Do I need to expand these powers of six?' The answer is a resounding no. In the world of JEE Advanced, we do not solve problems by brute force; we solve them by elegance and strategy.

Let us peel back the layers of this problem together.

The Divide and Conquer Strategy

Imagine you are a general on a battlefield. You do not attack the entire enemy army at once; you split them.

We have a product of two major components. Let us define:

Our expression is now $P(x)=(2-x^2)\cdot [A(x)+B(x)]$. By using the distributive property, we can rewrite this as:

This is the turning point. We have transformed one massive, intimidating expression into two distinct, manageable tasks. We are no longer looking for the coefficient of $x^2$ in the whole thing; we are looking for it in two specific, smaller pieces.

The Hunt for Coefficients

Now, let us analyze what we actually need. In the first part, $2\cdot [A(x)+B(x)]$, we need the coefficient of $x^2$. This means we need to find the $x^2$ coefficient within the sum $A(x)+B(x)$ and multiply it by two.

In the second part, $-x^2\cdot [A(x)+B(x)]$, we are already multiplying by $x^2$. This is a beautiful shortcut! If we multiply $x^2$ by any term higher than a constant, we get $x^3$ or $x^4$, which we do not care about.

Therefore, we only need the constant term of $A(x)+B(x)$. Our master plan is set:

The Multinomial Challenge

Let us tackle $A(x)=(1+2x+3x^2{)}^6$. This is a trinomial raised to a power. The Binomial Theorem cannot help us here, so we invoke the Multinomial Theorem.

The general term is given by: where $p+q+r=6$. The power of $x$ is $q+2r$. We need this power to be 2. So, we set $q+2r=2$.

Since $p,q,r$ must be non-negative integers, we have two cases:

Case 1: If $r=1$, then $q=0$. This forces $p=5$. The term becomes:

Case 2: If $r=0$, then $q=2$. This forces $p=4$. The term becomes:

Adding these, the coefficient of $x^2$ in $A(x)$ is $18+60=78$.

The Binomial Simplicity

Now for $B(x)=(1-4x^2{)}^6$. This is a simple binomial. We need the $x^2$ term.

Using the expansion $\left(\genfrac{}{}{0ex}{}{6}{0}\right)+\left(\genfrac{}{}{0ex}{}{6}{1}\right)(-4x^2{)}^1+\dots$, the $x^2$ term is clearly:

The constant term is simply $\left(\genfrac{}{}{0ex}{}{6}{0}\right)=1$. It is that simple!

The Final Assembly

We are at the finish line. The total coefficient of $x^2$ in $A(x)+B(x)$ is $78+(-24)=54$. The total constant term is $1+1=2$.

Plugging these into our master plan:

Look at that! We navigated the complexity, broke it down, and arrived at the solution with precision. This is the power of mathematical thinking. You didn't just solve a problem; you mastered a technique. The final answer is 106.