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Q.
Consider a function $f : N \rightarrow R$, 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
Given for $x \geq 2$
$ f(1)+2 f(2)+\ldots \ldots+x f(x)=x(x+1) f(x)$
$\text { replace } x \text { by } x+1 $
$\Rightarrow x(x+1) f(x)+(x+1) f(x+1)$
$ =( x +1)( x +2) f ( x +1)$
$ \Rightarrow \frac{x}{f(x+1)}+\frac{1}{f(x)}=\frac{(x+2)}{f(x)}$
$ \Rightarrow x f(x)=(x+1) f(x+1)=\frac{1}{2}, x \geq 2$
$ f (2)=\frac{1}{4}, f (3)=\frac{1}{6}$
$ \text { Now } f(2022)=\frac{1}{4044}$
$f(2028)=\frac{1}{4056} $
$ \text { So, } \frac{1}{f(2022)}+\frac{1}{f(2028)}=4044+4056=8100 $