Problem Analysis #5

[THOUGHT] The function $f_n(x)$ is defined as: $f_n(x) := \sum_{k=1}^{n^2} 123456 \left(\frac{x}{2}\right)^{k-1} \left(1 - \frac{x}{2}\right)^3$ First, let's simplify the summand: $123456 \left(\frac{x}{2}\right)^{k-1} \left(1 - \frac{x}{2}\right)^3$ We can rewrite it as: $123456 \left(\frac{x}{2}\right)^{k-1} \left(1 - \frac{x}{2}\right)^3 = 123456 \left(\frac{x}{2}\right)^{k-1} \left(1 - \frac{x}{2}\right)^3$ This is a weighted sum of terms involving a power of $\frac{x}{2}$ and a constant factor $\left(1 - \frac{x}{2}\right)^3$.

This expression resembles a finite geometric series, but with a fixed exponent on the second term. Let's consider the behavior of $f_n(x)$ as $n \to \infty$.

For $x = 0$: $f_n(0) = \sum_{k=1}^{n^2} 123456 \left(0\right)^{k-1} \left(1 - 0\right)^3 = 123456 \cdot n^2$

For $0 < x < 2$: $f_n(x) = 123456 \left(1 - \frac{x}{2}\right)^3 \sum_{k=1}^{n^2} \left(\frac{x}{2}\right)^{k-1}$ The sum inside is a geometric series: $\sum_{k=1}^{n^2} \left(\frac{x}{2}\right)^{k-1} = \frac{1 - \left(\frac{x}{2}\right)^{n^2}}{1 - \frac{x}{2}}$ Thus, $f_n(x) = 123456 \left(1 - \frac{x}{2}\right)^2 \frac{1 - \left(\frac{x}{2}\right)^{n^2}}{\frac{2-x}{2}} = \frac{123456 \cdot 2 \left(1 - \frac{x}{2}\right)^2}{2-x} \left(1 - \left(\frac{x}{2}\right)^{n^2}\right)$ $f_n(x) = \frac{2471280 \left(1 - \frac{x}{2}\right)^2}{2-x} \left(1 - \left(\frac{x}{2}\right)^{n^2}\right)$

As $n \to \infty$, $\left(\frac{x}{2}\right)^{n^2} \to 0$ for $0 < x < 2$, so: $\lim_{n \to \infty} f_n(x) = \frac{2471280 \left(1 - \frac{x}{2}\right)^2}{2-x}$

Now, let's integrate this limit function over $[0, 2]$: $I = \int_0^2 \frac{2471280 \left(1 - \frac{x}{2}\right)^2}{2-x} \, dx$

To simplify the integrand, let's substitute $u = \frac{x}{2}$, hence $du = \frac{1}{2} dx$ or $dx = 2 du$. When $x = 0$, $u = 0$, and when $x = 2$, $u = 1$. The integral becomes: $I = \int_0^1 \frac{2471280 \left(1 - u\right)^2}{2-2u} \cdot 2 \, du = \int_0^1 \frac{2471280 \cdot 2 (1-u)^2}{2(1-u)} \, du$ $I = \int_0^1 2471280 (1-u) \, du$ $I = 2471280 \int_0^1 (1-u) \, du$ $I = 2471280 \left[ u - \frac{u^2}{2} \right]_0^1$ $I = 2471280 \left( 1 - \frac{1}{2} \right)$ $I = 2471280 \cdot \frac{1}{2}$ $I = 1235640$

Therefore, the value of the limit is: $\boxed{1235640}$ [/THOUGHT]