$f : \mathbb{N} \to \mathbb{R}$ satisfies $f(1) + 2f(2) + \dots + xf(x) = x(x + 1)f(x)$ for $x \geq 2$, with $f(1) = 1$. Then $\frac{1}{f(2022)} + \frac{1}{f(2023)}$ equals:
Step-by-Step Solution
Key Concept: Write the relation for $x$ and for $x - 1$, then subtract to find a recurrence for $x \cdot f(x)$. Show $x \cdot f(x)$ is constant, find $f(x)$, then evaluate.
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Correct Answer: (2)