The value of $$k \in \mathbb{N}$$ for which the integral $$I_n = \int_0^1 (1 - x^k)^n dx, n \in \mathbb{N}$$, satisfies $$147I_{20} = 148I_{21}$$ is

We consider $$I_n = \int_0^1 (1-x^k)^n \, dx$$ for natural numbers $$n$$ and $$k$$, with the condition $$147 I_{20} = 148 I_{21}$$.

Since substituting $$t = x^k$$ so that $$x = t^{1/k}$$ and $$dx = \frac{1}{k} t^{1/k - 1} \, dt$$, it follows that $$I_n = \int_0^1 (1-t)^n \cdot \frac{1}{k} t^{1/k - 1} \, dt = \frac{1}{k} B\left(\frac{1}{k}, n+1\right) = \frac{1}{k} \cdot \frac{\Gamma(1/k) \cdot \Gamma(n+1)}{\Gamma(n + 1 + 1/k)}.$$

This gives the ratio $$\frac{I_n}{I_{n+1}} = \frac{\Gamma(n+1) \cdot \Gamma(n+2+1/k)}{\Gamma(n+2) \cdot \Gamma(n+1+1/k)} = \frac{n+1+1/k}{n+1} = 1 + \frac{1}{k(n+1)}.$$

From the condition $$147 I_{20} = 148 I_{21}$$ we obtain $$\frac{I_{20}}{I_{21}} = \frac{148}{147}.$$ Substituting $$n = 20$$ in the ratio formula yields $$\frac{I_{20}}{I_{21}} = 1 + \frac{1}{21k} = \frac{21k + 1}{21k}.$$

Equating these expressions gives $$\frac{21k + 1}{21k} = \frac{148}{147},$$ $$147(21k + 1) = 148 \times 21k,$$ $$147 \times 21k + 147 = 148 \times 21k,$$ $$147 = 21k,$$ and hence $$k = 7.$$

The answer is Option 4: 7.