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A function f: N -> R satisfies f(1) = 1 and the relation f(1) + 2f(2) + 3f(3) +... + x*f(x) = x(x+1)*f(x) for all integers x >= 2. Find the value of 1/f(2022) + 1/f(2028).

1. 8200
2. 8000
3. 8400
4. 8100

Correct answer: 8100

Solution

Subtracting consecutive cases of the given relation yields x*f(x) = (x-1)*f(x-1) for x >= 3. Plugging x=2 directly gives f(2) = 1/4, so the constant value of x*f(x) is 1/2, giving f(x) = 1/(2x) for x >= 2. Thus 1/f(2022) + 1/f(2028) = 2*2022 + 2*2028 = 4044 + 4056 = 8100.

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