<p>The number of terms in the expansion of \(\left(1 - \dfrac{2}{x} + \dfrac{4}{x^2}\right)^n\) is \({}^{n+2}C_2 = 28\). Find the sum of coefficients of the expansion.</p>
Step-by-Step Solution
Key Concept: Recognize that (1 - 2/x + 4/x²)ⁿ = [(1 - 2/x + 4/x²)]ⁿ can be rewritten as (x² - 2x + 4)ⁿ/x^(2n), so the number of distinct terms depends on the degrees of x in the expansion. The number of terms in a trinomial expansion (a + b + c)ⁿ is (n+1)(n+2)/2 = ⁿ⁺²C₂, which gives us n directly.
<p><strong>Step 1:</strong> Use the formula for number of terms in (a + b + c)ⁿ expansion, which is ⁿ⁺²C₂ = (n+2)(n+1)/2 = 28</p><p><strong>Step 2:</strong> Solve: (n+2)(n+1)/2 = 28 → (n+2)(n+1) = 56 → n² + 3n + 2 = 56 → n² + 3n - 54 = 0</p><p><strong>Step 3:</strong> Factor: (n + 9)(n - 6) = 0 → n = 6 (taking positive value)</p><p><strong>Step 4:</strong> Find sum of coefficients by substituting x = 1 in the original expression: (1 - 2(1) + 4(1)²)⁶ = (1 - 2 + 4)⁶ = 3⁶ = 729</p><p>∴ Answer: A (729)</p>
Correct Answer: A