The smallest natural number n, such that the coefficient of x in the expansion of `(x^(2) + (1)/(x^(3)))^(n)` is `.^(n)C_(23)`, is

Options

A. 58
B. 38
C. 23
D. 35

Explanation

To solve the problem, we need to find the smallest natural number \( n \) such that the coefficient of \( x \) in the expansion of \( (x^2 + \frac{1}{x^3})^n \) is equal to \( \binom{n}{23} \).

### Step-by-step Solution:

1. **Identify the General Term**:
The general term \( T_{r+1} \) in the expansion of \( (x^2 + \frac{1}{x^3})^n \) can be expressed as:
\[ T_{r+1} = \binom{n}{r} (x^2)^{n-r} \left(\frac{1}{x^3}\right)^r = \binom{n}{r} x^{2(n-r)} \cdot x^{-3r} = \binom{n}{r} x^{2n - 5r} \]

2. **Set the Power of \( x \) to 1**:
To find the coefficient of \( x \), we need the exponent of \( x \) to be 1:
\[ 2n - 5r = 1 \]
Rearranging gives:
\[ 5r = 2n - 1 \quad \Rightarrow \quad r = \frac{2n - 1}{5} \]

3. **Find the Coefficient**:
The coefficient of \( x \) in this term is given by:
\[ \binom{n}{r} = \binom{n}{\frac{2n - 1}{5}} \]
According to the problem, this coefficient must equal \( \binom{n}{23} \):
\[ \binom{n}{\frac{2n - 1}{5}} = \binom{n}{23} \]

4. **Equate the Indices**:
Since \( \binom{n}{k} = \binom{n}{n-k} \), we can set:
\[ \frac{2n - 1}{5} = 23 \quad \text{or} \quad \frac{2n - 1}{5} = n - 23 \]

5. **Solve for \( n \)**:
**Case 1**: Solving \( \frac{2n - 1}{5} = 23 \):
\[ 2n - 1 = 115 \quad \Rightarrow \quad 2n = 116 \quad \Rightarrow \quad n = 58 \]

**Case 2**: Solving \( \frac{2n - 1}{5} = n - 23 \):
\[ 2n - 1 = 5(n - 23) \quad \Rightarrow \quad 2n - 1 = 5n - 115 \quad \Rightarrow \quad 3n = 114 \quad \Rightarrow \quad n = 38 \]

6. **Determine the Smallest Natural Number**:
The possible values of \( n \) from both cases are \( 58 \) and \( 38 \). The smallest natural number \( n \) is:
\[ n = 38 \]

### Final Answer:
The smallest natural number \( n \) is \( \boxed{38} \).