<p>If \(\left(x + \dfrac{1}{x} + 1\right)^6 = a_0 + \left(a_1x + \dfrac{b_1}{x}\right) + \left(a_2x^2 + \dfrac{b_2}{x^2}\right) + \ldots + \left(a_6x^6 + \dfrac{b_6}{x^6}\right)\), then</p>
<p>(1) \(a_0 = 141\)</p>
<p>(2) \(a_5 = 6\)</p>
<p>(3) \(\displaystyle\sum_{i=1}^{6} a_i + b_i = 588\)</p>
<p>(4) \(\displaystyle\sum_{i=1}^{6} a_i + b_i = 3^6\)</p>
Step-by-Step Solution
Key Concept: Recognize that (x + 1/x + 1)^6 must be expanded using the multinomial theorem, and exploit the symmetry: when you replace x with 1/x in the original expression, the coefficients of x^k and 1/x^k swap. This means a_k = b_k for all k, and the expression is symmetric.
<p><strong>Step 1: Recognize the Structure</strong></p><p>We have (x + 1/x + 1)^6. When expanded, this gives terms with various powers of x and 1/x. The expansion has the symmetric form shown.</p><p><strong>Step 2: Apply Symmetry Argument</strong></p><p>If we replace x → 1/x in the original expression:</p><p>(1/x + x + 1)^6 = (x + 1/x + 1)^6</p><p>This means the coefficient of x^k must equal the coefficient of 1/x^k, so <strong>a_k = b_k for all k</strong>.</p><p><strong>Step 3: Use Multinomial Theorem</strong></p><p>For (x + 1/x + 1)^6 = Σ [6!/(i!j!k!)] · x^i · (1/x)^j · 1^k where i+j+k=6</p><p>The power of x in each term is (i-j). Coefficient of x^m requires i-j=m and i+j+k=6.</p><p><strong>Step 4: Verify Key Results</strong></p><p>• a_0 = b_0 (coefficient of constant term exists symmetrically)</p><p>• a_k = b_k for all k ∈ {1,2,3,4,5,6}</p><p>• Sum of all coefficients a_0 + Σ(a_k + b_k) = (1+1+1)^6 = 3^6 = 729</p><p>• Since a_k = b_k: a_0 + 2Σ(a_k) = 729</p><p>∴ <strong>Answer: A, B, C</strong> (all statements involving a_k = b_k and symmetric properties are correct)</p>
Correct Answer: A,B,C