<p><strong>235.</strong> 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>(a) \(3^{10} \cdot {}^{19}C_8\)</p>
<p>(b) \(3^{10} \cdot {}^{19}C_9\)</p>
<p>(c) \(3^9 \cdot {}^{20}C_8\)</p>
<p>(d) \(3^9 \cdot {}^{19}C_{40}\)</p>
Step-by-Step Solution
Key Concept: For a term to be independent of x in binomial expansion, the power of x must equal zero. Set up the general term and equate the exponent of x to 0, then use the binomial coefficient formula.
<p><strong>Step 1: Find A (independent term in first expansion)</strong></p><p>General term: T_{r+1} = C(20,r)(3x)^{20-r}(-1/x)^r = C(20,r)·3^{20-r}·(-1)^r·x^{20-2r}</p><p>For independence: 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>Rewrite: √[9]{3^{10}} = 3^{10/9}</p><p>General term: T_{s+1} = C(18,s)·x^{18-s}·(3^{10/9}/x)^s = C(18,s)·3^{10s/9}·x^{18-2s}</p><p>For independence: 18 - 2s = 0 ⟹ s = 9</p><p>B = C(18,9)·3^{10·9/9} = C(18,9)·3^{10}</p><p><strong>Step 3: Calculate the final expression</strong></p><p>Note: C(20,10) = 184756 and C(18,9) = 48620</p><p>(9/38)A + B = (9/38)·C(20,10)·3^{10} + C(18,9)·3^{10}</p><p>= 3^{10}[(9/38)·184756 + 48620]</p><p>= 3^{10}[43758 + 48620]</p><p>= 3^{10}·92378</p><p>∴ Answer: B</p>
Correct Answer: B