<p>Find the coefficient of \(x^7\) in the expansion of \((1 + 3x - 2x^3)^{10}\).</p>
Step-by-Step Solution
Key Concept: Use the multinomial theorem to expand (1 + 3x - 2x³)¹⁰ by considering all ways to select terms from each factor such that the powers of x sum to 7. This requires finding non-negative integers a, b, c where a + b + c = 10 and b + 3c = 7.
<p><strong>Step 1:</strong> In the expansion of (1 + 3x - 2x³)¹⁰, the general term is:</p><p>$$\frac{10!}{a!b!c!} \cdot 1^a \cdot (3x)^b \cdot (-2x^3)^c = \frac{10!}{a!b!c!} \cdot 3^b \cdot (-2)^c \cdot x^{b+3c}$$</p><p>where a + b + c = 10.</p><p><strong>Step 2:</strong> For coefficient of x⁷, we need b + 3c = 7 with a + b + c = 10.</p><p>From b + 3c = 7: b = 7 - 3c</p><p>From a + b + c = 10: a = 10 - b - c = 10 - (7 - 3c) - c = 3 + 2c</p><p>For non-negative integers: c ∈ {0, 1, 2} (since b ≥ 0 requires c ≤ 2⅓)</p><p><strong>Step 3:</strong> Calculate each case:</p><p><strong>Case 1:</strong> c = 0 → a = 3, b = 7, c = 0<br/>Coefficient: $$\frac{10!}{3!7!0!} \cdot 3^7 \cdot (-2)^0 = 120 \cdot 2187 = 262440$$</p><p><strong>Case 2:</strong> c = 1 → a = 5, b = 4, c = 1<br/>Coefficient: $$\frac{10!}{5!4!1!} \cdot 3^4 \cdot (-2)^1 = 1260 \cdot 81 \cdot (-2) = -204120$$</p><p><strong>Case 3:</strong> c = 2 → a = 7, b = 1, c = 2<br/>Coefficient: $$\frac{10!}{7!1!2!} \cdot 3^1 \cdot (-2)^2 = 360 \cdot 3 \cdot 4 = 4320$$</p><p><strong>Step 4:</strong> Total coefficient = 262440 - 204120 + 4320 = <strong>62640</strong></p><p><em>Note: If the answer provided (−173556) is correct, verify the problem statement. The calculation above is standard for this problem type.</em></p>
Correct Answer: -173556