Animated Solution for Mathematics - Binomial Theorem: If the coefficients of $x^3$ and $x^4$ in the expansion of $(1+ax+b{x}^2)(1-2x{)}^{18}$ in powers of $x$ are both zero, then $(a,b)$ is equal to

Imagine you are standing before a massive, daunting expression: $(1+ax+b{x}^2)(1-2x{)}^{18}$. At first glance, it looks like a nightmare of algebra.

If you were to try and expand this fully, you would be writing for hours, drowning in a sea of terms. But here is the secret of the JEE Advanced: we don't need the whole expansion.

We only need the coefficients of $x^3$ and $x^4$. We are not here to do brute-force labor; we are here to be surgeons.

When we multiply two polynomials, say $P(x)$ and $Q(x)$, the coefficient of $x^n$ in the product is not just one term. It is a symphony of interactions.

Every term in $P(x)$ reaches out to a term in $Q(x)$ such that their powers sum to $n$. For our expression, $P(x)=1+ax+b{x}^2$ and $Q(x)=(1-2x{)}^{18}$.

To find the coefficient of $x^3$, we look at the components of $P(x)$: 1. The constant $1$ must multiply the $x^3$ term of $Q(x)$. 2. The term $ax$ must multiply the $x^2$ term of $Q(x)$. 3. The term $b{x}^2$ must multiply the $x^1$ term of $Q(x)$.

Using the Binomial Theorem, the general term of $(1-2x{)}^{18}$ is $T_{r+1}=\left(\genfrac{}{}{0ex}{}{18}{r}\right)(-2{)}^r{x}^r$. This is our toolkit. By plugging in $r=1,2,3,$ and $4$, we can extract exactly what we need without ever touching the rest of the expansion.

Taming $x^3$

Let us assemble our first equation. The coefficient of $x^3$ is the sum of our three interactions:

Calculating these values, we get $-8(816)+4a(153)-2b(18)=0$. This simplifies beautifully to $-6528+612a-36b=0$.

If we divide by 12, we arrive at a clean, elegant linear equation: $51a-3b=544$. This is our first anchor point.

The $x^4$ Challenge

Now, we repeat the process for $x^4$. The logic remains identical, but the stakes are slightly higher. We need the coefficient of $x^4$ to be zero:

Substituting the values, we get $16(3060)-8a(816)+4b(153)=0$.

After dividing by 4 and then by 51, we find the second equation: $32a-3b=240$.

We now have a system of two linear equations with two variables. It is a moment of pure algebraic satisfaction.

By subtracting the equations, the $b$ terms vanish, leaving us with $19a=304$, which yields $a=16$.

Substituting this back into our equations, we find $b=\frac{272}{3}$.

We have successfully navigated the complexity. We didn't need to expand the 18th power; we simply understood the structure of the product. Remember, in mathematics, the most powerful tool is not the one that does the most work, but the one that reveals the underlying pattern.