<p>If \(x > 0\), the first negative term in the expansion of \((1+x)^{27/5}\) is:</p>
Step-by-Step Solution
Key Concept: Use the general term formula T_{r+1} = C(27/5, r)·x^r and find when the binomial coefficient becomes negative by analyzing the sign of the product (27/5)(27/5 - 1)(27/5 - 2)...(27/5 - r + 1). The coefficient is negative when the numerator of the product crosses zero.
<p><strong>Step 1:</strong> For the expansion of (1+x)^(27/5), the general term is:</p><p>T_{r+1} = C(27/5, r)·x^r where C(27/5, r) = [(27/5)(27/5 - 1)(27/5 - 2)...(27/5 - r + 1)]/r!</p><p><strong>Step 2:</strong> Analyze the numerator product (27/5)(22/5)(17/5)(12/5)(7/5)(2/5)(-3/5)(-8/5)...</p><p>We have: 27/5 > 0, 22/5 > 0, 17/5 > 0, 12/5 > 0, 7/5 > 0, 2/5 > 0, -3/5 < 0</p><p><strong>Step 3:</strong> The first negative factor appears at (27/5 - 6) = -3/5, which corresponds to r = 7.</p><p>For r = 0 to 6: product has 7 positive terms → coefficient is positive</p><p>For r = 7: product has 6 positive and 1 negative term → coefficient is negative</p><p><strong>Step 4:</strong> Since x > 0, the term T_8 = C(27/5, 7)·x^7 is the first negative term (as C(27/5, 7) < 0 and x^7 > 0).</p><p>∴ Answer: The 8th term (or T_8)</p>
Correct Answer: D