Solution: Finding Coefficients in a Binomial Expansion

1. Key Concept: Binomial Theorem and Polynomial Multiplication

The problem requires us to find the coefficients of specific powers of $x$ in the product of two polynomials: a quadratic $(1 + ax + bx^2)$ and a binomial raised to a power $(1 - 2x)^{18}$.

The key tool here is the Binomial Theorem, which states that for any non-negative integer $n$, the expansion of $(p+q)^n$ is given by: $(p+q)^n = \sum_{r=0}^{n} \binom{n}{r} p^{n-r} q^r$ where $\binom{n}{r} = \frac{n!}{r!(n-r)!}$ is the binomial coefficient.

In our case, for the term $(1 - 2x)^{18}$, we have $p=1$, $q=-2x$, and $n=18$. The general term $T_{r+1}$ (the $(r+1)^{th}$ term) in this expansion is: $T_{r+1} = \binom{18}{r} (1)^{18-r} (-2x)^r = \binom{18}{r} (-2)^r x^r$ Let $C_r = \binom{18}{r} (-2)^r$ be the coefficient of $x^r$ in the expansion of $(1 - 2x)^{18}$.

The given expression is $P(x) = (1 + ax + bx^2) (1 - 2x)^{18}$. To find the coefficient of $x^k$ in $P(x)$, we need to identify all combinations of terms from $(1 + ax + bx^2)$ and $(1 - 2x)^{18}$ that multiply to give $x^k$.

2. Step-by-Step Working

Step 2.1: Calculate relevant binomial coefficients and powers of $(-2)$

We need coefficients up to $x^4$. Let's compute the necessary binomial coefficients $\binom{18}{r}$ and powers of $(-2)$:

• $C_0 = \binom{18}{0} (-2)^0 = 1 \times 1 = 1$
• $C_1 = \binom{18}{1} (-2)^1 = 18 \times (-2) = -36$
• $C_2 = \binom{18}{2} (-2)^2 = \frac{18 \times 17}{2 \times 1} \times 4 = 9 \times 17 \times 4 = 153 \times 4 = 612$
• $C_3 = \binom{18}{3} (-2)^3 = \frac{18 \times 17 \times 16}{3 \times 2 \times 1} \times (-8) = 3 \times 17 \times 16 \times (-8) = 816 \times (-8) = -6528$
• $C_4 = \binom{18}{4} (-2)^4 = \frac{18 \times 17 \times 16 \times 15}{4 \times 3 \times 2 \times 1} \times 16 = 18 \times 17 \times 10 \times 16 = 3060 \times 16 = 48960$

So, the expansion of $(1 - 2x)^{18}$ starts as $1 - 36x + 612x^2 - 6528x^3 + 48960x^4 + \dots$

Step 2.2: Determine the coefficient of $x^3$

The terms contributing to $x^3$ in the expansion of $(1 + ax + bx^2) (C_0 + C_1 x + C_2 x^2 + C_3 x^3 + C_4 x^4 + \dots)$ are:

• $1 \times (C_3 x^3)$
• $ax \times (C_2 x^2)$
• $bx^2 \times (C_1 x)$

Summing these up, the coefficient of $x^3$ is $C_3 + aC_2 + bC_1$. According to the problem, this coefficient is zero. $C_3 + aC_2 + bC_1 = 0$ $-6528 + a(612) + b(-36) = 0$ $-6528 + 612a - 36b = 0$ To simplify, divide the entire equation by $12$: $-544 + 51a - 3b = 0$ $51a - 3b = 544 \quad \text{(Equation 1)}$

Step 2.3: Determine the coefficient of $x^4$

Similarly, the terms contributing to $x^4$ are:

• $1 \times (C_4 x^4)$
• $ax \times (C_3 x^3)$
• $bx^2 \times (C_2 x^2)$

Summing these up, the coefficient of $x^4$ is $C_4 + aC_3 + bC_2$. According to the problem, this coefficient is also zero. $C_4 + aC_3 + bC_2 = 0$ $48960 + a(-6528) + b(612) = 0$ $48960 - 6528a + 612b = 0$ To simplify, divide the entire equation by $12$: $4080 - 544a + 51b = 0$ $544a - 51b = 4080 \quad \text{(Equation 2)}$

Step 2.4: Solve the system of linear equations

We now have a system of two linear equations with two variables $a$ and $b$:

1. $51a - 3b = 544$
2. $544a - 51b = 4080$

From Equation 1, we can express $3b$ in terms of $a$: $3b = 51a - 544$ $b = \frac{51a - 544}{3}$

Substitute this expression for $b$ into Equation 2: $544a - 51 \left( \frac{51a - 544}{3} \right) = 4080$ Simplify the term with $51$: $51/3 = 17$. $544a - 17(51a - 544) = 4080$ $544a - (17 \times 51)a + (17 \times 544) = 4080$ $544a - 867a + 9248 = 4080$ $-323a = 4080 - 9248$ $-323a = -5168$ $a = \frac{-5168}{-323}$ $a = 16$

Now, substitute the value of $a=16$ back into the simplified Equation 1 ($51a - 3b = 544$): $51(16) - 3b = 544$ $816 - 3b = 544$ $3b = 816 - 544$ $3b = 272$ $b = \frac{272}{3}$

Thus, the values are $a=16$ and $b=\frac{272}{3}$. The pair $(a, b)$ is $\left(16, \frac{272}{3}\right)$.

3. Tips and Common Mistakes to Avoid

• Sign Errors: Be extremely careful with the negative sign in $(-2x)$. Forgetting it or incorrectly applying it for even and odd powers is a common mistake. $(-2x)^r = (-2)^r x^r$.
• Missing Terms: Ensure you account for all combinations that produce the desired power of $x$. For example, for $x^k$, remember to consider $1 \cdot C_k x^k$, $ax \cdot C_{k-1} x^{k-1}$, and $bx^2 \cdot C_{k-2} x^{k-2}$.
• Calculation Errors: Binomial coefficients and multiplications can involve large numbers. Double-check your arithmetic, especially when simplifying equations. Using a calculator for intermediate steps can be helpful, but understanding the process is paramount.
• Simplifying Equations: Divide by common factors as early as possible to keep numbers manageable, reducing the chance of error during system solving.
• Options Discrepancy: In competitive exams, sometimes there might be slight errors in the provided options or the "correct answer." Always trust your systematic calculation. If your result doesn't directly match an option, re-check your steps. If the discrepancy persists, choose the option closest to your calculated value or highlight the potential error if possible. In this specific problem, our derived answer $(16, \frac{272}{3})$ matches option (B), not the provided correct answer (A).

4. Summary and Key Takeaway

This problem demonstrates a systematic approach to finding coefficients in polynomial products, especially when one factor is a binomial expansion. The core idea is to:

1. Expand the binomial term using the Binomial Theorem to find its general terms.
2. Identify all combinations of terms from both polynomial factors that yield the desired power of $x$.
3. Formulate linear equations by setting these combined coefficients to zero (as specified in the problem).
4. Solve the resulting system of equations to find the unknown parameters.

Mastering the binomial theorem and careful algebraic manipulation are crucial for solving such problems efficiently and accurately.