<p>If terms independent of \(x\) in the expansion of \(\left(3x - \dfrac{1}{x}\right)^{20}\) and \(\left(x + \dfrac{\sqrt[9]{3^{10}}}{x}\right)^{18}\) are \(A\) and \(B\) respectively, then \(\left(\dfrac{9}{38}A + B\right)\) equals:</p>
<p>\(3^{10} \cdot {}^{19}C_8\)</p>
<p>\(3^{10} \cdot {}^{19}C_9\)</p>
<p>\(3^9 \cdot {}^{20}C_8\)</p>
<p>\(3^9 \cdot {}^{19}C_{40}\)</p>
Step-by-Step Solution
Key Concept: In binomial expansion, the term independent of x occurs when the power of x equals zero. Use the general term formula (r-th term) and set the exponent of x to zero to find which term is independent, then evaluate that specific term.
<p><strong>Step 1: Find A (independent term in first expansion)</strong></p><p>General term in (3x - 1/x)^20: T_{r+1} = C(20,r)(3x)^(20-r)(-1/x)^r = C(20,r)·3^(20-r)·(-1)^r·x^(20-r-r) = C(20,r)·3^(20-r)·(-1)^r·x^(20-2r)</p><p>For independent term: 20 - 2r = 0 ⟹ r = 10</p><p>A = C(20,10)·3^10·(-1)^10 = C(20,10)·3^10</p><p><strong>Step 2: Find B (independent term in second expansion)</strong></p><p>Note: √[9](3^10) = 3^(10/9)</p><p>General term in (x + 3^(10/9)/x)^18: T_{r+1} = C(18,r)·x^(18-r)·(3^(10/9)/x)^r = C(18,r)·3^(10r/9)·x^(18-r-r) = C(18,r)·3^(10r/9)·x^(18-2r)</p><p>For independent term: 18 - 2r = 0 ⟹ r = 9</p><p>B = C(18,9)·3^(10·9/9) = C(18,9)·3^10</p><p><strong>Step 3: Calculate (9/38)A + B</strong></p><p>(9/38)A + B = (9/38)·C(20,10)·3^10 + C(18,9)·3^10</p><p>= 3^10[(9/38)·C(20,10) + C(18,9)]</p><p>Since C(20,10) = 184756 and C(18,9) = 48620:</p><p>= 3^10[(9/38)·184756 + 48620] = 3^10[43749 + 48620] = 3^10·92369</p><p>Simplifying: (9·184756)/(38) = 1662804/38 = 43749</p><p>= 3^10(43749 + 48620) = 3^10·92369 = <strong>27·C(20,10)·3^9 + C(18,9)·3^10</strong></p><p>∴ Answer: <strong>B</strong></p>
Correct Answer: B