Key Concepts and Formulas

• Summation Properties: If $S_x = \sum_{i=1}^x i f(i)$, then $S_x - S_{x-1} = x f(x)$ for $x \ge 2$.
• Recurrence Relations: A formula that defines a sequence recursively, where each term is defined as a function of preceding terms.
• Solving Linear First-Order Recurrence Relations: Techniques to find a closed-form expression for the terms of a sequence defined by a recurrence relation.

Step-by-Step Solution

Step 1: Define the Summation and Rewrite the Given Relation Let $S_x = \sum_{i=1}^x i f(i)$. The given relation is $S_x = x(x+1)f(x)$ for $x \ge 2$.

Step 2: Express $x f(x)$ in terms of the Summation From the definition of $S_x$, we can write $S_x = S_{x-1} + x f(x)$ for $x \ge 2$. Therefore, $x f(x) = S_x - S_{x-1}$.

Step 3: Substitute the Given Relation into the Expression for $x f(x)$ We are given $S_x = x(x+1)f(x)$ for $x \ge 2$. Also, for $x-1 \ge 2$ (i.e., $x \ge 3$), we have $S_{x-1} = (x-1)((x-1)+1)f(x-1) = (x-1)x f(x-1)$.

Now, substitute these into the equation from Step 2: $x f(x) = x(x+1)f(x) - (x-1)x f(x-1)$ for $x \ge 3$.

Step 4: Simplify the Equation to Obtain a Recurrence Relation for $f(x)$ Divide the entire equation by $x$ (since $x \ge 3$, $x \neq 0$): $f(x) = (x+1)f(x) - (x-1)f(x-1)$ for $x \ge 3$.

Rearrange the terms to isolate $f(x)$: $(x+1)f(x) - f(x) = (x-1)f(x-1)$ $x f(x) = (x-1)f(x-1)$ for $x \ge 3$.

This simplifies to: $f(x) = \frac{x-1}{x} f(x-1)$ for $x \ge 3$.

Step 5: Handle the Case for $x=2$ The given relation $S_x = x(x+1)f(x)$ holds for $x \ge 2$. For $x=2$: $S_2 = 2(2+1)f(2) = 6f(2)$. Also, $S_2 = 1 \cdot f(1) + 2 \cdot f(2)$. We are given $f(1) = 1$. So, $S_2 = 1 + 2f(2)$.

Equating the two expressions for $S_2$: $6f(2) = 1 + 2f(2)$ $4f(2) = 1$ $f(2) = \frac{1}{4}$.

Let's check if the recurrence $f(x) = \frac{x-1}{x} f(x-1)$ holds for $x=2$ if we use $f(1)$. If we formally apply it for $x=2$: $f(2) = \frac{2-1}{2} f(1) = \frac{1}{2} f(1) = \frac{1}{2} \cdot 1 = \frac{1}{2}$. This is not equal to $f(2) = 1/4$. This indicates that the recurrence relation derived in Step 4 is valid only for $x \ge 3$.

Step 6: Find a General Formula for $f(x)$ for $x \ge 3$ We have the recurrence $f(x) = \frac{x-1}{x} f(x-1)$ for $x \ge 3$. Let's expand this: $f(x) = \frac{x-1}{x} f(x-1)$ $f(x-1) = \frac{x-2}{x-1} f(x-2)$ $f(x-2) = \frac{x-3}{x-2} f(x-3)$ ... $f(3) = \frac{2}{3} f(2)$

Substituting these back: $f(x) = \frac{x-1}{x} \cdot \frac{x-2}{x-1} \cdot \frac{x-3}{x-2} \cdots \frac{2}{3} f(2)$ The intermediate terms cancel out (telescoping product). $f(x) = \frac{2}{x} f(2)$ for $x \ge 3$.

We found $f(2) = 1/4$. So, for $x \ge 3$, $f(x) = \frac{2}{x} \cdot \frac{1}{4} = \frac{1}{2x}$.

Let's verify this formula for $x=3$: $f(3) = \frac{1}{2 \cdot 3} = \frac{1}{6}$. Using the recurrence $f(3) = \frac{3-1}{3} f(2) = \frac{2}{3} f(2) = \frac{2}{3} \cdot \frac{1}{4} = \frac{2}{12} = \frac{1}{6}$. This matches.

Step 7: Check the Formula for $f(1)$ and $f(2)$ Our formula $f(x) = \frac{1}{2x}$ is derived for $x \ge 3$. For $x=1$: $f(1) = 1$. The formula gives $1/(2 \cdot 1) = 1/2$, which is incorrect. For $x=2$: $f(2) = 1/4$. The formula gives $1/(2 \cdot 2) = 1/4$. So, the formula $f(x) = \frac{1}{2x}$ works for $x \ge 2$.

Thus, we have: $f(1) = 1$ $f(x) = \frac{1}{2x}$ for $x \ge 2$.

Step 8: Calculate the Required Terms We need to find $\frac{1}{f(2022)} + \frac{1}{f(2028)}$. Since $2022 \ge 2$ and $2028 \ge 2$, we use the formula $f(x) = \frac{1}{2x}$.

$\frac{1}{f(2022)} = \frac{1}{1/(2 \cdot 2022)} = 2 \cdot 2022 = 4044$. $\frac{1}{f(2028)} = \frac{1}{1/(2 \cdot 2028)} = 2 \cdot 2028 = 4056$.

Step 9: Compute the Final Sum $\frac{1}{f(2022)} + \frac{1}{f(2028)} = 4044 + 4056 = 8100$.

Common Mistakes & Tips

• Incorrectly applying the recurrence: The recurrence relation $f(x) = \frac{x-1}{x} f(x-1)$ was derived for $x \ge 3$. It's crucial to check if it holds for the boundary cases ($x=2$ in this instance) or handle them separately.
• Telescoping product errors: When simplifying the product for $f(x)$, ensure all intermediate terms cancel correctly and the correct base case value ($f(2)$ in this case) is used.
• Forgetting the initial condition: The value of $f(1)$ is essential for determining the specific form of the function, especially if the derived formula doesn't hold for $x=1$.

Summary

The problem provides a functional equation involving a summation. By defining the summation $S_x$ and using the property $S_x - S_{x-1} = x f(x)$, we rewrote the given equation to derive a recurrence relation for $f(x)$. We carefully handled the case for $x=2$ separately, as the general recurrence was valid for $x \ge 3$. Solving the recurrence relation and using the value of $f(2)$ led to the general form $f(x) = \frac{1}{2x}$ for $x \ge 2$, with $f(1)=1$. Finally, we used this formula to calculate the required sum $\frac{1}{f(2022)} + \frac{1}{f(2028)}$.

The final answer is \boxed{8100}.