Consider a function [f:NR], satisfying [f(1)+2f(2)+3f(3)+....+xf(x)=x(x+1)f(x);x2] with f(…

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

Accepted Answer

$$f(1) + 2f(2) + 3f(3)\, + \,...\, + \,nf(n) = n(n + 1) + (n)$$ ..... (i)

$$n \to n + 1$$

$$f(1) + 2f(2)\, + \,...\, + \,(n + 1)f(n + 1) = (n + 1)(n + 2)f(n + 1)$$ ...... (ii)

(i) and (ii) gives

$3f(3) - 2f(2) = 0$

$4f(4) - 3f(3) = 0$

$$ \vdots $$

$(n + 1)f(n + 1) - nf(n) = 0$

$$ \Rightarrow f(n + 1) = {{2f(2)} \over {n + 1}}$$

$$f(n) = {1 \over {2n}}$$

$${1 \over {f(2022)}} + {1 \over {f(2028)}} = 8100$$

Solutions · 1