[THOUGHT] Let me analyze this limit. We have: $\lim_{n\to\infty} \int_{[0,n]} \left(1+\frac{205}{69n}x\right)^{-69n} e^{-\frac{6}{7}x}\,dx$

First, I need to find the pointwise limit of the integrand as $n \to \infty$.

For the term $\left(1+\frac{205}{69n}x\right)^{-69n}$, let me use the fact that $\lim_{n\to\infty}\left(1+\frac{a}{n}\right)^n = e^a$.

So $\left(1+\frac{205x}{69n}\right)^{-69n} = \left[\left(1+\frac{205x}{69n}\right)^{69n}\right]^{-1} \to e^{-205x}$ as $n \to \infty$.

Therefore, the integrand converges pointwise to $e^{-205x} \cdot e^{-\frac{6}{7}x} = e^{-x(205 + \frac{6}{7})} = e^{-x \cdot \frac{1441}{7}}$.

By dominated convergence (the integrand is bounded by something integrable), the limit should be: $\int_0^\infty e^{-\frac{1441}{7}x}\,dx = \frac{7}{1441}$

Let me verify this calculation and check if 7 and 1441 are coprime. [/THOUGHT] [PYTHON] import math from fractions import Fraction

Calculate 205 + 6/7

exponent_coeff = 205 + Fraction(6, 7) print(f"Exponent coefficient: {exponent_coeff}")

The integral of e^(-ax) from 0 to infinity is 1/a

result = 1 / exponent_coeff print(f"Result: {result}")

Get p and q

p = result.numerator q = result.denominator print(f"p = {p}, q = {q}")

Check if they are coprime

gcd = math.gcd(p, q) print(f"GCD(p, q) = {gcd}")

Calculate p^2 * q

answer = p**2 * q print(f"p^2 * q = {answer}") [/PYTHON]

[THOUGHT] Let me verify the calculation:

1. The integrand is $\left(1+\frac{205}{69n}x\right)^{-69n} e^{-\frac{6}{7}x}$

2. As $n \to \infty$, we have $\left(1+\frac{205x}{69n}\right)^{-69n} \to e^{-205x}$ (using the standard limit $\lim_{n\to\infty}(1+a/n)^n = e^a$)

3. So the integrand converges to $e^{-205x} \cdot e^{-\frac{6}{7}x} = e^{-x(205 + \frac{6}{7})} = e^{-\frac{1441x}{7}}$

4. The integral $\int_0^\infty e^{-\frac{1441x}{7}} dx = \frac{7}{1441}$

5. Since $\gcd(7, 1441) = 1$, we have $p = 7$ and $q = 1441$

6. Therefore, $p^2 q = 49 \times 1441 = 70609$ [/THOUGHT]

\boxed{70609}