If the 4th term in the expansion of $$\left( mx + \frac{1}{x} \right)^n$$is  $$\frac{5}{2}$$ then what is the value of mn?

Correct option is B

Solution:

$$\text{The binomial expansion is:} \quad \left( mx + \frac{1}{x} \right)^n\\\textbf{Step 1: Identify the 4th Term}\\\text{The general term in the binomial expansion of } (a + b)^n \text{ is given by:}\\\quad T_{k+1} = \binom{n}{k} a^{n-k} b^k\\\text{For the given expression, } a = mx, \, b = \frac{1}{x}, \text{ and the 4th term corresponds to } k = 3 \, (\text{since terms are counted from } k = 0):\\\quad T_4 = \binom{n}{3} (mx)^{n-3} \left( \frac{1}{x} \right)^3\\\textbf{Step 2: Simplify the 4th Term}\\\text{Simplify the expression for } T_4:\\\quad T_4 = \binom{n}{3} m^{n-3} x^{n-3} \cdot x^{-3}\\\quad T_4 = \binom{n}{3} m^{n-3} x^{n-6}\\\text{The problem states that } T_4 = \frac{5}{2}. \\\text{ However, } T_4 \text{ is expressed in terms of } x, \text{ which suggests that the coefficient of } x^{n-6} \text{ is } \frac{5}{2}. \\\text{ Therefore:}\quad \binom{n}{3} m^{n-3} = \frac{5}{2}\\\text{Additionally, for the term to be independent of } x \text{ (as } \frac{5}{2} \text{ is a constant), the exponent of } x \text{ must be zero:}\quad n - 6 = 0 \quad \Rightarrow \quad n = 6\\\textbf{Step 3: Substitute } n = 6 \\\text{ and Solve for } m\\\text{Substitute } n = 6 \text{ into the equation for the coefficient:}\quad \binom{6}{3} m^{6-3} = \frac{5}{2} \quad \Rightarrow \quad \binom{6}{3} m^3 = \frac{5}{2}\quad \Rightarrow \quad 20 m^3 = \frac{5}{2}\\\text{Solve for } m:\quad m^3 = \frac{5}{2} \div 20 = \frac{5}{40} \quad \Rightarrow \quad m = \frac{1}{2}\\\textbf{Step 4: Compute } mn\text{Now, multiply } m \text{ and } n:\\\quad mn = \frac{1}{2} \times 6 = 3$$​