JEE Math Practice Question
Consider a function $$f:\mathbb{N}\to\mathbb{R}$$, satisfying $$f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x\ge2$$ with $$f(1)=1$$. Then $$\frac{1}{f(2022)}+\frac{1}{f(2028)}$$ is equal to
- A.8000
- B.8400
- C.8100
- D.8200
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