Binomial Expansion 4th Term Calculation

The problem asks us to find the value of the product \(mn\) based on the information about the \(4\)th term in the binomial expansion of \(\left(mx + \frac{1}{x}\right)^n\). The \(4\)th term is given as \(\frac{5}{2}\).

Understanding the Binomial Expansion Formula

The formula for the general term (the \((r+1)^{th}\) term) in the binomial expansion of \((a+b)^n\) is given by:

\(T_{r+1} = \binom{n}{r} a^{n-r} b^r\)

In this specific expansion, \(\left(mx + \frac{1}{x}\right)^n\), we have:

• \(a = mx\)
• \(b = \frac{1}{x}\)
• \(n\) is the exponent

Calculating the 4th Term

We are interested in the \(4\)th term, which means \(r+1 = 4\), so \(r=3\). Plugging the values into the general term formula:

\(T_4 = T_{3+1} = \binom{n}{3} (mx)^{n-3} \left(\frac{1}{x}\right)^3\)

Let's simplify this expression:

\(T_4 = \binom{n}{3} m^{n-3} x^{n-3} \cdot \frac{1^3}{x^3}\)

\(T_4 = \binom{n}{3} m^{n-3} x^{n-3} \cdot \frac{1}{x^3}\)

Combine the powers of \(x\):

\(T_4 = \binom{n}{3} m^{n-3} x^{(n-3)-3}\)

\(T_4 = \binom{n}{3} m^{n-3} x^{n-6}\)

Determining the Value of n

The problem states that the \(4\)th term is a constant value, \(\frac{5}{2}\). For the term \(T_4 = \binom{n}{3} m^{n-3} x^{n-6}\) to be a constant, the power of \(x\) must be zero. This is because \(m\) and \(n\) are constants, and \(\binom{n}{3}\) is a constant value.

Therefore, we set the exponent of \(x\) to zero:

\(n-6 = 0\)

Solving for \(n\) gives:

\(n = 6\)

Solving for the Value of m

Now that we know \(n=6\), we can substitute this back into the expression for \(T_4\):

\(T_4 = \binom{6}{3} m^{6-3} x^{6-6}\)

\(T_4 = \binom{6}{3} m^3 x^0\)

Calculate the binomial coefficient \(\binom{6}{3}\):

\(\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20\)

So, the expression for \(T_4\) becomes:

\(T_4 = 20 \times m^3 \times 1\)

\(T_4 = 20 m^3\)

We are given that the \(4\)th term is \(\frac{5}{2}\). Equating this with our calculated term:

\(20 m^3 = \frac{5}{2}\)

Now, solve for \(m^3\):

\(m^3 = \frac{5}{2 \times 20}\)

\(m^3 = \frac{5}{40}\)

\(m^3 = \frac{1}{8}\)

Taking the cube root of both sides to find \(m\):

\(m = \sqrt[3]{\frac{1}{8}}\)

\(m = \frac{1}{2}\)

Final Calculation of mn

We have found the values of \(m\) and \(n\):

• \(m = \frac{1}{2}\)
• \(n = 6\)

The question asks for the value of \(mn\). Let's calculate it:

\(mn = \left(\frac{1}{2}\right) \times 6\)

\(mn = 3\)

Thus, the value of \(mn\) is \(3\). This corresponds to the second option.