Given:
f(1)+2f(2)+3f(3)+…+xf(x)=x(x+1)f(x)
⇒f(1)+2f(2)+3f(3)+…+(x−1)f(x−1)=x(x−1)f(x−1)
Now when x=2, then
f(1)+2f(2)=6f(2)
⇒1=4f(2)
⇒f(2)=41
When x=3,
f(1)+2f(2)=9f(3)⇒f(3)=61
When x=4, f(1)+2f(2)+3f(3)=16f(4)⇒f(4)=81
So,
f(x)=2x1
Hence,
f(2022)1+f(2028)1=4044+4056=8100