JEE MAIN - Mathematics (2023 - 29th January Evening Shift - No. 7)
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
8000
8400
8100
8200
Explanation
$$f(1) + 2f(2) + 3f(3)\, + \,...\, + \,nf(n) = n(n + 1) + (n)$$ ..... (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
$$3f(3) - 2f(2) = 0$$
$$4f(4) - 3f(3) = 0$$
$$ \vdots $$
$$(n + 1)f(n + 1) - nf(n) = 0$$
$$ \Rightarrow f(n + 1) = {{2f(2)} \over {n + 1}}$$
$$f(n) = {1 \over {2n}}$$
$${1 \over {f(2022)}} + {1 \over {f(2028)}} = 8100$$
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