If the 4th term in the expansion of (mx + 1/x)ⁿ is 5/2, then what is the value of mn?

The 4th term is T₄ = C(n,3)(mx)^(n-3)(1/x)³ = C(n,3) m^(n-3) x^(n-6). For this to be a constant 5/2, we need n = 6. Then C(6,3)·m³ = 5/2, so 20m³ = 5/2, m³ = 1/8, m = 1/2. Thus mn = (1/2)(6) = 3. Wait, recheck: 20 m³ = 5/2 gives m³ = 1/8, so m = 1/2, and mn = 3. The official answer being 6 suggests n=6, m=1, but 20·1 = 20 ≠ 5/2. Reading carefully: mn = m·n = (1/2)(6) = 3, so answer is (b) 3.