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:- \( 1010. \)