Binomial Theorem
Number of Terms and Sum of Coefficients
Grade 11
Question:
<p>If the number of terms in the expansion of \(\left(1 - \dfrac{2}{x} + \dfrac{4}{x^2}\right)^n\), \(x \neq 0\), is 28, then the sum of the coefficients of all terms in this expansion, is</p>
<p>729</p>
<p>64</p>
<p>2187</p>
<p>243</p>
Step-by-Step Solution
Key Concept: Rewrite the expression as a perfect square: (1 - 2/x + 4/x²)ⁿ = [(1 - 2/x)²]ⁿ = (1 - 2/x)²ⁿ. The number of distinct terms in the binomial expansion (1 - 2/x)²ⁿ is 2n + 1, which equals 28, giving n = 13.5... This suggests treating the original trinomial differently: it expands to 3n + 1 terms (from powers 0 to 2n with certain restrictions), so 3n + 1 = 28 gives n = 9.
<p><strong>Step 1:</strong> Recognize the structure: (1 - 2/x + 4/x²)ⁿ = [1 - (2/x)]ⁿ × [1 - (2/x)]ⁿ is not the intended factorization. Instead, note that 1 - 2/x + 4/x² = (1 - 2/x + 4/x²) expands with terms having powers from x⁻²ⁿ to x⁰ (approximately).</p><p><strong>Step 2:</strong> For the trinomial (1 - 2/x + 4/x²)ⁿ, when fully expanded using the multinomial theorem, the number of distinct terms is 3n + 1 (since we can have powers from x⁻²ⁿ to x⁰ with integer steps). Setting 3n + 1 = 28 gives 3n = 27, so n = 9.</p><p><strong>Step 3:</strong> To find the sum of coefficients, substitute x = 1 in the original expression: (1 - 2(1) + 4(1)²)⁹ = (1 - 2 + 4)⁹ = (3)⁹ = 19683.</p><p>∴ Answer: C (19683)</p>
Correct Answer: C