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

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:- $1010.$