Consider a function $\mathrm { f } : \mathbb { N } \rightarrow \mathbb { R }$, satisfying $f ( 1 ) + 2 f ( 2 ) + 3 f ( 3 ) + \ldots + x f ( x ) = x ( x + 1 ) f ( x ) ; x \geq 2$ with $f ( 1 ) = 1$. Then $\frac { 1 } { f ( 2022 ) } + \frac { 1 } { f ( 2028 ) }$ is equal to
(1) 8200
(2) 8000
(3) 8400
(4) 8100