Consider a function f:\mathbb{N} o\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

Question

Consider a function f:\mathbb{N}	o\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

Accepted Answer

Answer: C. 8100 f(1) + 2f(2) + 3f(3)\, + \,...\, + \,nf(n) = n(n + 1) + (n) ..... (i) n 	o n + 1 f(1) + 2f(2)\, + \,...\, + \,(n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1) ...... (ii) (i) and (ii) gives 3f(3) - 2f(2) = 0 4f(4) - 3f(3) = 0 \vdots (n + 1)f(n + 1) - nf(n) = 0 \Rightarrow f(n + 1) = {{2f(2)} \over {n + 1}} f(n) = {1 \over {2n}} {1 \over {f(2022)}} + {1 \over {f(2028)}} = 8100

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

Answer

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