Question

The coefficient of \( x^3 \) in the expansion of \( (1 + x + 2x^2)(1 - 2x)^5 \) is

• A. \(-20\)
• B. \(-40\)
• C. \(-60\)
• D. \(-80\)
• E. \(-100\)

Hint:

Break multiplication into contributions from each term to avoid missing cases.

Answer 1

The Correct Option is C

Concept: To find the coefficient of a specific power (here \(x^3\)), we expand using the binomial theorem and carefully collect only those terms whose total power of \(x\) becomes 3.

Step 1: Expand \( (1-2x)^5 \) using binomial theorem. General term: \[ T_k = \binom{5}{k}(-2x)^k = \binom{5}{k}(-2)^k x^k \]

Step 2: Identify contributions to \(x^3\). We multiply: \[ (1 + x + 2x^2)(\text{terms of } (1-2x)^5) \] We need combinations giving total power \(x^3\):
• \(1 \times (\text{term in } x^3)\)
• \(x \times (\text{term in } x^2)\)
• \(2x^2 \times (\text{term in } x^1)\)

Step 3: Find each required coefficient. (i) Coefficient of \(x^3\) in \( (1-2x)^5 \): \[ \binom{5}{3}(-2)^3 = 10 \times (-8) = -80 \] (ii) Coefficient of \(x^2\): \[ \binom{5}{2}(-2)^2 = 10 \times 4 = 40 \] Contribution: \[ x \cdot 40x^2 = 40x^3 \Rightarrow 40 \] (iii) Coefficient of \(x\): \[ \binom{5}{1}(-2)^1 = 5 \times (-2) = -10 \] Contribution: \[ 2x^2 \cdot (-10x) = -20x^3 \Rightarrow -20 \]

Step 4: Add all contributions. \[ -80 + 40 - 20 = -60 \]