Consider a function \( f:\mathbb{N} \rightarrow \mathbb{R} \), satisfying \( f(1) + 2f(2) + 3f(3) + \ldots + x f(x) = {x(x+1)}{}; \, x \ge 2 \) with \( f(1) = 1 \). Then \( \frac{1}{f(2022)} + \frac{1}{f(2028)} \) is equal to:
\( 8000 \)
\( 8200 \)
\( 8400 \)
\( 8100 \)