[THOUGHT] To solve this limit problem, I need to evaluate the integral and then take the limit as ( n ) approaches infinity. The integrand resembles the form of an exponential function multiplied by a term that looks like a limit definition of the exponential function. I recall that: [ \lim_{n \to \infty} \left(1 + \frac{a}{n}x\right)^n = e^{a x}. ] Using this, the term (\left(1 + \frac{205}{69n}x\right)^{-69n}) can be rewritten as: [ \left(1 + \frac{205}{69n}x\right)^{-69n} = \left[\left(1 + \frac{205}{69n}x\right)^{\frac{69n}{205x}}\right]^{-\frac{205x}{69}} \to e^{-\frac{205x}{69}} \text{ as } n \to \infty. ] Thus, the integrand converges to ( e^{-\frac{205x}{69}} \cdot e^{-\frac{6}{7}x} = e^{-\left(\frac{205}{69} + \frac{6}{7}\right)x} ).

Now, I need to compute the integral of ( e^{-\left(\frac{205}{69} + \frac{6}{7}\right)x} ) from 0 to (\infty). The integral of ( e^{-k x} ) from 0 to (\infty) is ( \frac{1}{k} ), provided ( k > 0 ).

First, I will compute the coefficient ( k = \frac{205}{69} + \frac{6}{7} ). Then, the limit of the integral will be ( \frac{1}{k} ). Finally, I will express ( \frac{1}{k} ) in reduced form ( \frac{p}{q} ) and compute ( p^2 q ).

[/THOUGHT] [PYTHON] from fractions import Fraction

Compute the coefficient k

k = Fraction(205, 69) + Fraction(6, 7)

Simplify k

k_simplified = Fraction(k).limit_denominator()

The integral evaluates to 1/k

p_over_q = Fraction(1, k_simplified)

Ensure p and q are coprime

p = p_over_q.numerator q = p_over_q.denominator

Compute p^2 q

result = p**2 * q print(result) [/PYTHON]

[THOUGHT] The Python output confirms that ( p^2 q = 431351361 ). This is the final answer. [/THOUGHT]

\boxed{431351361}