Question

If a function $f$ satisfies $f(m + n) = f(m) + f(n)$ for all $m, n \in \mathbb{N}$ and $f(1) = 1$, then the largest natural number $\lambda$ such that $$\sum_{k=1}^{2022} f(\lambda + k) \leq (2022)^2$$ is equal to __________.

Answer 1

Correct Answer: 1010

Step 1: Analyze the functional equation The functional equation $f(m+n) = f(m) + f(n)$ suggests that $f(x)$ is linear. Assume:

$$f(x) = kx.$$

Substitute $f(1) = 1$:

$$f(1) = k \cdot 1 \implies k = 1.$$

Thus:

$$f(x) = x.$$

Step 2: Expand the summation We are given:

$$\sum_{k=1}^{2022} f(\lambda + k) \leq (2022)^2.$$

Substitute $f(x) = x$:

$$\sum_{k=1}^{2022} (\lambda + k) \leq (2022)^2.$$

Split the summation:

$$\sum_{k=1}^{2022} (\lambda + k) = \sum_{k=1}^{2022} \lambda + \sum_{k=1}^{2022} k.$$

Step 2.1: Simplify each term

1. $$\sum_{k=1}^{2022} \lambda = 2022 \cdot \lambda.$$
2. $$\sum_{k=1}^{2022} k = \frac{2022 \cdot 2023}{2} \quad \text{(sum of the first 2022 natural numbers)}.$$

Thus:

$$\sum_{k=1}^{2022} (\lambda + k) = 2022 \cdot \lambda + \frac{2022 \cdot 2023}{2}.$$

Step 3: Solve the inequality Substitute into the inequality:

$$2022\lambda + \frac{2022 \cdot 2023}{2} \leq (2022)^2.$$

Simplify:

$$2022\lambda \leq (2022)^2 - \frac{2022 \cdot 2023}{2}.$$

Factor 2022 out:

$$2022\lambda \leq 2022 \left( 2022 - \frac{2023}{2} \right).$$

Simplify further:

$$\lambda \leq 2022 - \frac{2023}{2}.$$

Calculate:

$$\lambda \leq 2022 - 1011.5 = 1010.5.$$

Step 4: Largest natural number Since $\lambda$ must be a natural number:

$$\lambda = 1010.$$

Final Answer:- $1010.$