The smallest natural number n, such that the coefficient of x in the expansion of $$\left(x^2 + \frac{1}{x^3}\right)^n$$ is $$^nC_{23}$$, is

We need to find the smallest natural number $$n$$ such that the coefficient of $$x$$ in the expansion of $$\left(x^2 + \frac{1}{x^3}\right)^n$$ is $$^nC_{23}$$.

The general term in the binomial expansion of $$\left(x^2 + \frac{1}{x^3}\right)^n$$ is:

$$T_{r+1} = \binom{n}{r} (x^2)^{n-r} \left(\frac{1}{x^3}\right)^r = \binom{n}{r} x^{2(n-r)} \cdot x^{-3r} = \binom{n}{r} x^{2n - 2r - 3r} = \binom{n}{r} x^{2n - 5r}$$

For the coefficient of $$x$$, we need the power of $$x$$ to be 1:

$$2n - 5r = 1$$ $$-(1)$$

This gives $$r = \frac{2n-1}{5}$$. For $$r$$ to be a non-negative integer, $$2n - 1$$ must be divisible by 5.

The coefficient of $$x$$ in the expansion is $$\binom{n}{r}$$, and we are told this equals $$\binom{n}{23}$$:

$$\binom{n}{r} = \binom{n}{23}$$

Using the property of binomial coefficients, $$\binom{n}{r} = \binom{n}{23}$$ implies either:

Case 1: $$r = 23$$, or

Case 2: $$r = n - 23$$ (using the symmetry property $$\binom{n}{k} = \binom{n}{n-k}$$)

Case 1: If $$r = 23$$, substituting into equation $$(1)$$:

$$2n - 5(23) = 1$$

$$2n = 116$$

$$n = 58$$

Case 2: If $$r = n - 23$$, substituting into equation $$(1)$$:

$$2n - 5(n-23) = 1$$

$$2n - 5n + 115 = 1$$

$$-3n = -114$$

$$n = 38$$

We need to verify that $$r$$ is a valid non-negative integer in each case:

For $$n = 38$$: $$r = 38 - 23 = 15$$, and $$0 \leq 15 \leq 38$$ ✔

For $$n = 58$$: $$r = 23$$, and $$0 \leq 23 \leq 58$$ ✔

The smallest value of $$n$$ is $$38$$.

The correct answer is Option B: 38.