Problem Analysis #4

[THOUGHT] Let $f(x) = \left(1+\frac{205}{69n}x\right)^{-69n} e^{(-\frac{6}{7}x)}$. As $n \to \infty$, we have $\frac{205}{69n} \to 0$. Thus, [ \left(1+\frac{205}{69n}x\right)^{-69n} \approx e^{-\frac{205}{69n}x \cdot 69n} = e^{-205x}. ] Therefore, the integral becomes [ \lim_{n\to\infty} \int_{[0,n]} e^{-205x-\frac{6}{7}x},dx = \lim_{n\to\infty} \int_0^n e^{-(205+\frac{6}{7})x},dx. ] We have $205 + \frac{6}{7} = \frac{1435+6}{7} = \frac{1441}{7}$. [ \lim_{n\to\infty} \int_0^n e^{-\frac{1441}{7}x},dx = \lim_{n\to\infty} \left[-\frac{7}{1441}e^{-\frac{1441}{7}x}\right]0^n = \lim{n\to\infty} -\frac{7}{1441}e^{-\frac{1441}{7}n} - \left(-\frac{7}{1441}e^0\right) = \lim_{n\to\infty} -\frac{7}{1441}e^{-\frac{1441}{7}n} + \frac{7}{1441}. ] Since $\lim_{n\to\infty} e^{-\frac{1441}{7}n} = 0$, we have [ \lim_{n\to\infty} \int_0^n e^{-(205+\frac{6}{7})x},dx = \frac{7}{1441}. ] Thus, $p=7$ and $q=1441$. We need to check if they are relatively prime. $1441 = 7 \cdot 205 + 6$. Thus $\gcd(7,1441) = \gcd(7,6)=1$. Then $p^2q = 7^2 \cdot 1441 = 49 \cdot 1441 = 70609$.

p = 7
q = 1441
print(p**2 * q)

[THOUGHT] The Python output confirms the calculation $7^2 \cdot 1441 = 70609$.

\boxed{70609}