<p>Given \(\sum_{r=0}^{2n} a_r \left(\frac{1}{x}\right)^r = \left(1 + \frac{1}{x} + \frac{1}{x^2}\right)^n\) and \(a_r = a_{2n-r}\) for \(0 \le r \le n-1\), find the value of \(2(a_0 + a_1 + \ldots + a_{n-1}) + a_n\).</p>
Step-by-Step Solution
Key Concept: The expansion (1 + 1/x + 1/x²)ⁿ produces coefficients with symmetry property aᵣ = a₂ₙ₋ᵣ. Use the substitution x=1 to find the sum of all coefficients, then exploit symmetry to isolate the desired expression.
<p><strong>Step 1:</strong> Expand (1 + 1/x + 1/x²)ⁿ using the multinomial theorem. The general coefficient aᵣ depends on how powers of 1/x combine.</p><p><strong>Step 2:</strong> Set x = 1 in the original equation: Σ aᵣ = (1 + 1 + 1)ⁿ = 3ⁿ. This gives a₀ + a₁ + ... + a₂ₙ = 3ⁿ.</p><p><strong>Step 3:</strong> Use the symmetry property aᵣ = a₂ₙ₋ᵣ. For r = 0,1,...,n-1, we have aᵣ = a₂ₙ₋ᵣ where indices 2n-r correspond to r = n+1, n+2, ..., 2n.</p><p><strong>Step 4:</strong> Therefore: a₀ + a₁ + ... + aₙ₋₁ = aₙ₊₁ + aₙ₊₂ + ... + a₂ₙ</p><p><strong>Step 5:</strong> From Step 2: (a₀ + a₁ + ... + aₙ₋₁) + aₙ + (aₙ₊₁ + ... + a₂ₙ) = 3ⁿ</p><p><strong>Step 6:</strong> Substituting the symmetry result: 2(a₀ + a₁ + ... + aₙ₋₁) + aₙ = 3ⁿ</p><p>∴ Answer: <strong>B</strong> (3ⁿ)</p>
Correct Answer: B