Binomial Theorem
Coefficient in multinomial expansion
Grade 11

Question:

<p>The coefficient of \(x^7\) in the expansion of \((1 - x - x^2 + x^3)^6\) is</p>
<p>132</p>
<p>144</p>
<p>\(-132\)</p>
<p>\(-144\)</p>

Step-by-Step Solution

Key Concept: Factor the expression as (1-x)(1-x²)³ × (1-x)³ = (1-x)⁴(1-x²)³, then use binomial expansion to find the coefficient of x⁷ by finding all pairs of powers that sum to 7.
<p><strong>Step 1:</strong> Factor the base expression.</p><p>1 - x - x² + x³ = (1 - x) - x²(1 - x) = (1 - x)(1 - x²)</p><p><strong>Step 2:</strong> Rewrite the expansion.</p><p>(1 - x - x² + x³)⁶ = [(1-x)(1-x²)]⁶ = (1-x)⁶(1-x²)⁶</p><p><strong>Step 3:</strong> Express in binomial form.</p><p>(1-x)⁶ = Σ C(6,r)(-x)ʳ and (1-x²)⁶ = Σ C(6,s)(-x²)ˢ</p><p><strong>Step 4:</strong> Find coefficient of x⁷.</p><p>We need r + 2s = 7, where 0 ≤ r ≤ 6 and 0 ≤ s ≤ 6.</p><p>Valid pairs: (r=7, s=0) - invalid; (r=5, s=1); (r=3, s=2); (r=1, s=3)</p><p><strong>Step 5:</strong> Calculate contribution from each pair.</p><p>From (r=5, s=1): C(6,5)(-1)⁵ · C(6,1)(-1)¹ = (6)(-1)(6)(-1) = 36</p><p>From (r=3, s=2): C(6,3)(-1)³ · C(6,2)(-1)² = (20)(-1)(15)(1) = -300</p><p>From (r=1, s=3): C(6,1)(-1)¹ · C(6,3)(-1)³ = (6)(-1)(20)(-1) = 120</p><p><strong>Step 6:</strong> Sum all contributions.</p><p>Coefficient of x⁷ = 36 - 300 + 120 = -144</p><p>∴ Answer: D</p>
Correct Answer: D

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