Rewritten Solution

1. Key Concept: General Term of Binomial Expansion The problem asks us to find the coefficient of a specific power of $x$ in a binomial expansion. For any binomial expansion of the form $(a+b)^n$, the general term (or $(r+1)^{th}$ term), denoted as $T_{r+1}$, is given by the formula: $T_{r+1} = {}^nC_r \, a^{n-r} \, b^r$ where $r$ is a non-negative integer representing the term number (starting from $r=0$ for the first term), and $0 \le r \le n$.

2. Step-by-Step Working

Step 1: Identify the General Term for the given expansion. Our given expansion is ${\left( {{x^2} + {1 \over {{x^3}}}} \right)^n}$. Here, $a = x^2$ and $b = {1 \over {{x^3}}} = x^{-3}$. Substituting these into the general term formula: $T_{r+1} = {}^nC_r \, (x^2)^{n-r} \, (x^{-3})^r$ $T_{r+1} = {}^nC_r \, x^{2(n-r)} \, x^{-3r}$ $T_{r+1} = {}^nC_r \, x^{2n - 2r - 3r}$ $T_{r+1} = {}^nC_r \, x^{2n - 5r}$ Explanation: We use the power rules $(x^a)^b = x^{ab}$ and $x^a \cdot x^b = x^{a+b}$ to simplify the powers of $x$ and consolidate them into a single term.

Step 2: Determine the value of 'r' for the coefficient of x. We are looking for the coefficient of $x$ (which means $x^1$). Therefore, we need to set the exponent of $x$ in our general term equal to 1: $2n - 5r = 1$ Now, we solve for $r$ in terms of $n$: $5r = 2n - 1$ $r = \frac{2n - 1}{5}$ Explanation: By equating the exponent of $x$ in the general term to the desired exponent (which is 1 for $x^1$), we establish a relationship between $n$ and $r$. This value of $r$ will identify the specific term whose coefficient we need.

Step 3: Apply the condition for 'r'. For $r$ to be a valid term index in a binomial expansion, it must satisfy two conditions:

1. $r$ must be a non-negative integer ($r \ge 0$).
2. $r$ must be less than or equal to $n$ ($r \le n$). From $r = \frac{2n - 1}{5}$, we know that $2n-1$ must be a multiple of 5. Also, since $n$ is a natural number, $2n-1$ will always be positive, so $r \ge 0$ will be satisfied for $n \ge 1$. We will check $r \le n$ later.

Step 4: Equate the coefficient to the given value. The problem states that the coefficient of $x$ is ${}^nC_{23}$. From our general term, the coefficient of $x^{2n-5r}$ is ${}^nC_r$. So, we can write: ${}^nC_r = {}^nC_{23}$ Substituting the expression for $r$: ${}^nC_{\left( \frac{2n - 1}{5} \right)} = {}^nC_{23}$ Explanation: The combinatorial term ${}^nC_r$ represents the coefficient of $x^{2n-5r}$. Since we found the specific $r$ that gives $x^1$, we equate this coefficient to the value given in the problem.

Step 5: Solve for 'n' using the properties of combinations. A key property of combinations states that if ${}^nC_k = {}^nC_m$, then either $k = m$ or $k = n - m$. Applying this property to our equation: Let $k = \frac{2n - 1}{5}$ and $m = 23$.

Case 1: $k = m$ $\frac{2n - 1}{5} = 23$ $2n - 1 = 5 \times 23$ $2n - 1 = 115$ $2n = 116$ $n = 58$ Let's check if this value of $n$ yields a valid integer $r$: $r = \frac{2(58) - 1}{5} = \frac{116 - 1}{5} = \frac{115}{5} = 23$. This is a valid integer, and $0 \le 23 \le 58$, so this value of $n=58$ is a possible solution.

Case 2: $k = n - m$ $\frac{2n - 1}{5} = n - 23$ $2n - 1 = 5(n - 23)$ $2n - 1 = 5n - 115$ $115 - 1 = 5n - 2n$ $114 = 3n$ $n = \frac{114}{3}$ $n = 38$ Let's check if this value of $n$ yields a valid integer $r$: $r = \frac{2(38) - 1}{5} = \frac{76 - 1}{5} = \frac{75}{5} = 15$. This is a valid integer, and $0 \le 15 \le 38$, so this value of $n=38$ is also a possible solution. Explanation: We use the fundamental property of combinations to find all possible values of $n$. It's crucial to consider both possibilities ($k=m$ and $k=n-m$) to ensure no valid solutions are missed. After finding each potential $n$, we verify that the corresponding $r$ value is a non-negative integer and lies within the range $[0, n]$.

Step 6: Identify the smallest natural number 'n'. We found two possible values for $n$: $58$ and $38$. The question asks for the smallest natural number $n$. Comparing the two values, $38$ is smaller than $58$. Therefore, the smallest natural number $n$ is $38$.

3. Tips and Common Mistakes

• Don't forget the $nC_k = nC_{n-k}$ property: This is a very common oversight. Always consider both possibilities when solving ${}^nC_k = {}^nC_m$.
• Check 'r' conditions: Remember that $r$ must be a non-negative integer and $0 \le r \le n$. If any derived $r$ does not meet these criteria, the corresponding $n$ is invalid.
• Careful with exponent rules: Ensure you correctly combine the powers of $x$. A common mistake is $x^{2n-2r} \cdot x^{-3r} \ne x^{2n-2r+3r}$.

4. Summary To find the smallest natural number $n$ such that the coefficient of $x$ in the expansion of ${\left( {{x^2} + {1 \over {{x^3}}}} \right)^n}$ is ${}^nC_{23}$, we first determined the general term of the expansion. By setting the power of $x$ to $1$, we expressed $r$ in terms of $n$. Then, using the property ${}^nC_k = {}^nC_m \Rightarrow k=m$ or $k=n-m$, we found two possible values for $n$, which were $58$ and $38$. After verifying both solutions and considering the requirement for the smallest natural number, we concluded that $n=38$.

The final answer is $\boxed{\text{38}}$.