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

Explanation

The 4th term is T_4 = \binom{n}{3}(mx)^{n-3}(\frac{1}{x})^3 = \binom{n}{3}m^{n-3}x^{n-6}. For this to be a constant, the power of x must be zero, so n-6 = 0 \Rightarrow n = 6. Substituting n=6, T_4 = \binom{6}{3}m^3 = 20m^3. Given 20m^3 = \frac{5}{2} \Rightarrow m^3 = \frac{1}{8} \Rightarrow m = \frac{1}{2}. Therefore, mn = \frac{1}{2} \times 6 = 3.

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