<p>If \(x\) is positive, the first negative term in the expansion of \((1+x)^{27/5}\) is (\(|x| < 1\))</p>
Step-by-Step Solution
Key Concept: Use the general term formula for binomial expansion with fractional exponents: T_{r+1} = C(n,r)x^r where C(n,r) = n(n-1)(n-2)...(n-r+1)/r!. A term is negative when the coefficient becomes negative, which occurs when the numerator has an odd number of negative factors.
<p><strong>Step 1:</strong> Write the general term: 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> For r = 0,1,2,3,4,5: The numerator is (27/5)(22/5)(17/5)(12/5)(7/5)(2/5) — all factors are positive.</p><p><strong>Step 3:</strong> For r = 6: C(27/5, 6) = (27/5)(22/5)(17/5)(12/5)(7/5)(2/5)(-3/5)/6! — the factor (-3/5) appears, making this term negative.</p><p><strong>Step 4:</strong> Verify: Since 0 < x < 1, the x^6 factor is positive, so T_7 is the first negative term.</p><p>∴ Answer: D (The 7th term, or T_7)</p>
Correct Answer: D