[PYTHON] from sympy import symbols, integrate, limit, oo

x, n = symbols('x n') C = 123456

Define the term in the sum

sum_{k=1}^{n^2} (x/2)^(k-1)

This is a geometric series with ratio x/2, first term 1, number of terms n^2

Sum = (1 - (x/2)^(n^2)) / (1 - x/2)

f_n(x)

f_n = C * Sum * (1 - x/2)^3

f_n = C * (1 - (x/2)^(n^2)) / (1 - x/2) * (1 - x/2)^3

f_n = C * (1 - (x/2)^(n^2)) * (1 - x/2)^2

We want the limit of the integral of f_n from 0 to 2 as n -> oo

By DCT, this is the integral of the limit of f_n

limit of (x/2)^(n^2) as n->oo is 0 for x in [0, 2)

So limit function f(x) = C * (1 - 0) * (1 - x/2)^2 = C * (1 - x/2)^2

f_limit = C * (1 - x/2)**2 integral_val = integrate(f_limit, (x, 0, 2)) print(f"{integral_val=}") [/PYTHON]

The sequence of functions is given by [ f_n(x) = 123456 \left(1 - \frac{x}{2}\right)^3 \sum_{k=1}^{n^2} \left(\frac{x}{2}\right)^{k-1}. ] Let $r = \frac{x}{2}$. For $x \in [0, 2)$, we have $0 \le r < 1$. The sum is a geometric series: [ \sum_{k=1}^{n^2} r^{k-1} = \frac{1 - r^{n^2}}{1 - r}. ] Thus, for $x \in [0, 2)$, [ f_n(x) = 123456 (1 - r)^3 \frac{1 - r^{n^2}}{1 - r} = 123456 (1 - r)^2 (1 - r^{n^2}). ] As $n \to \infty$, $r^{n^2} \to 0$ for $r \in [0, 1)$. Therefore, the pointwise limit is [ f(x) = \lim_{n\to\infty} f_n(x) = 123456 \left(1 - \frac{x}{2}\right)^2. ] Since $|f_n(x)| \le 123456 (1 - x/2)^2$ on $[0, 2]$, and this dominating function is integrable, the Dominated Convergence Theorem applies: [ I = \int_0^2 f(x) , dx = \int_0^2 123456 \left(1 - \frac{x}{2}\right)^2 , dx. ] Let $u = 1 - \frac{x}{2}$, so $dx = -2 \, du$. The limits change from $[0, 2]$ to $[1, 0]$. [ I = 123456 \int_1^0 u^2 (-2) , du = 2 \cdot 123456 \int_0^1 u^2 , du = 246912 \left[ \frac{u^3}{3} \right]_0^1 = \frac{246912}{3} = 82304. ]

\boxed{82304}