If the fourth term in the expansion of `(px+1/x)^n` is independent of `x,` then the value of term is

Options

A. `5p^(3)`
B. `10p^(3)`
C. `20p^(3)`
D. None of these

Answer

To solve the problem, we need to find the value of \( n \) such that the fourth term in the expansion of \( (px + \frac{1}{x})^n \) is independent of \( x \).

### Step-by-Step Solution:

1. **Identify the General Term**:
The general term \( T_r \) in the expansion of \( (a + b)^n \) is given by:
\[ T_r = \binom{n}{r-1} a^{r-1} b^{n-(r-1)} \]
For our case, \( a = px \) and \( b = \frac{1}{x} \). Thus, we can write:
\[ T_r = \binom{n}{r-1} (px)^{r-1} \left(\frac{1}{x}\right)^{n-(r-1)} \]

2. **Write the Fourth Term**:
The fourth term \( T_4 \) corresponds to \( r = 4 \):
\[ T_4 = \binom{n}{3} (px)^3 \left(\frac{1}{x}\right)^{n-3} \]
Simplifying this, we get:
\[ T_4 = \binom{n}{3} p^3 x^3 \cdot \frac{1}{x^{n-3}} = \binom{n}{3} p^3 x^{3 - (n-3)} = \binom{n}{3} p^3 x^{6 - n} \]

3. **Condition for Independence of \( x \)**:
For the term to be independent of \( x \), the exponent of \( x \) must be zero:
\[ 6 - n = 0 \]
Solving this gives:
\[ n = 6 \]

4. **Calculate the Fourth Term**:
Now substituting \( n = 6 \) back into the expression for \( T_4 \):
\[ T_4 = \binom{6}{3} p^3 \]
We know that:
\[ \binom{6}{3} = \frac{6!}{3! \cdot 3!} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 \]
Therefore, the fourth term becomes:
\[ T_4 = 20 p^3 \]

### Final Answer:
The value of the fourth term is \( 20 p^3 \).