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Consider a function f:mathbbN→ mathbbR, satisfying f(1)+2f(2)+3f(3)+… +xf(x)=x(x+1)f(x);x≥ 2 with f(1)=1. Then 1 f(2022)+ 1 f(2028) is equal to

JEE Main 2023 — Mathematics Algebra

2023mcqmedium

Consider a function f:N→Rf:\mathbb{N}\rightarrow \mathbb{R}f:N→R, satisfying f(1)+2f(2)+3f(3)+…+xf(x)=x(x+1)f(x);x≥2f(1)+2f(2)+3f(3)+\ldots +xf(x)=x(x+1)f(x);x\geq 2f(1)+2f(2)+3f(3)+…+xf(x)=x(x+1)f(x);x≥2 with f(1)=1f(1)=1f(1)=1. Then 1f(2022)+1f(2028)\frac{1}{f(2022)}+\frac{1}{f(2028)}f(2022)1​+f(2028)1​ is equal to

Official previous-year question

Held on 29 Jan 2023 · Verified 6 Jul 2026.

Options

  1. A

    820082008200

  2. B

    800080008000

  3. C

    840084008400

  4. D

    810081008100

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