If a function $f$ satisfies $f(\mathrm{~m}+\mathrm{n})=f(\mathrm{~m})+f(\mathrm{n})$ for all $\mathrm{m}, \mathrm{n} \in \mathbf{N}$ and $f(1)=1$, then the largest natural number $\lambda$ such that $\sum_\limits{\mathrm{k}=1}^{2022} f(\lambda+\mathrm{k}) \leq(2022)^2$ is equal to _________.
Answer (integer)
1010
Solution
<p>$$\begin{aligned}
& f(m+n)=f(m)+f(n) \\
& f(x)=k x \\
& \because f(1)=1 \\
& \Rightarrow k=1 \\
& \Rightarrow f(x)=x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \sum_{k=1}^{2022} f(\lambda+k)=\sum_{k=1}^{2022}(\lambda+k)=\underbrace{\lambda+\lambda+\ldots+\lambda}_{2022}+(1+2+\ldots+2022) \\
& =2022 \lambda+\frac{2022 \times 2023}{2} \leq(2022)^2 \\
& \Rightarrow \lambda \leq \frac{2021}{2} \\
& \text { largest } \lambda=1010
\end{aligned}$$</p>
About this question
Subject: Mathematics · Chapter: Sets, Relations and Functions · Topic: Sets and Operations
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