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 __________.
Correct Answer: 1010
Solution and Explanation
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
- \[ \sum_{k=1}^{2022} \lambda = 2022 \cdot \lambda. \]
- \[ \sum_{k=1}^{2022} k = \frac{2022 \cdot 2023}{2} \quad \text{(sum of the first 2022 natural numbers)}. \]
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:- \( 1010. \)
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