JEE Mains 29-Jan-2023 Shift 2 Solved Paper

Section: Mathematics

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Question : 80 of 90

Marks: +1, -0

Consider a function f:ℕ→ℝ, 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

[29-Jan-2023 Shift 2]

8200
8000
8400
8100

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