Suppose the following limit is given:
where and are positive integers which are relatively prime. Find .
oindent
[THOUGHT] To solve this problem, I'll first simplify the integrand inside the limit. Note that as ( n \to \infty ), the expression (\left(1+\frac{205}{69n}x\right)^{-69n}) converges to the exponential function. Specifically, we recognize the standard limit:
[ \lim_{n \to \infty} \left(1 + \frac{a}{n}x\right)^{-n} = e^{-ax}. ]
In our problem, ( a = \frac{205}{69} ), so:
[ \left(1+\frac{205}{69n}x\right)^{-69n} \to e^{-\frac{205}{69}x}. ]
Therefore, the limit of the integrand becomes:
[ e^{-\frac{205}{69}x} \cdot e^{-\frac{6}{7}x} = e^{-\left(\frac{205}{69} + \frac{6}{7}\right)x}. ]
Next, we need to compute the sum:
[ \frac{205}{69} + \frac{6}{7}. ]
To add these fractions, find a common denominator, which is ( 483 ) (since ( 69 \times 7 = 483 )):
[ \frac{205}{69} = \frac{205 \times 7}{483} = \frac{1435}{483}, ] [ \frac{6}{7} = \frac{6 \times 69}{483} = \frac{414}{483}. ]
Now add the fractions:
[ \frac{1435}{483} + \frac{414}{483} = \frac{1849}{483}. ]
The integral then becomes:
[ \lim_{n\to\infty} \int_0^n e^{-\frac{1849}{483}x} , dx. ]
As ( n \to \infty ), this integral evaluates to the improper integral:
[ \int_0^\infty e^{-\frac{1849}{483}x} , dx. ]
This is a standard exponential integral which evaluates to:
[ \int_0^\infty e^{-kx} , dx = \frac{1}{k}, ]
where ( k = \frac{1849}{483} ). Thus, the integral is:
[ \frac{1}{\frac{1849}{483}} = \frac{483}{1849}. ]
We simplify this fraction by finding the greatest common divisor (GCD) of ( 483 ) and ( 1849 ). Using the Euclidean algorithm:
[ 1849 = 483 \times 3 + 420, ] [ 483 = 420 \times 1 + 63, ] [ 420 = 63 \times 6 + 42, ] [ 63 = 42 \times 1 + 21, ] [ 42 = 21 \times 2 + 0. ]
So, (\gcd(483, 1849) = 21). Simplifying:
[ \frac{483}{21} = 23, \quad \frac{1849}{21} = 89. ]
Thus, the fraction simplifies to (\frac{23}{89}), where 23 and 89 are coprime.
Finally, we need to find ( p^2q ) where ( \frac{p}{q} = \frac{23}{89} ):
[ p = 23, \quad q = 89. ]
So:
[ p^2q = 23^2 \times 89 = 529 \times 89. ]
I will now calculate ( 529 \times 89 ) using Python. [/THOUGHT] [PYTHON] p = 23 q = 89 result = p**2 * q print(result)
[THOUGHT] The calculation of ( 529 \times 89 ) using Python yielded the result 47081. Therefore, the value of ( p^2q ) is: [ \boxed{47081} ]