If a function $f$ satisfies $f(m+n)=f(m)+f(n)$ for all $m, n \in \mathbf{N}$ and $f(1)=1$, then the

Question

If a function $f$ satisfies $f(m+n)=f(m)+f(n)$ for all $m, n \in \mathbf{N}$ 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 $\_\_\_\_$ .\

Professor PC Thomas Classes · Answer

(1010): We have, $f(m+n)=f(m)+f(n)$ So, $f(x)=k x$\ $\because \quad f(1)=1 \Rightarrow k=1$\ Hence, $f(x)=x$\ Now, $\sum_{k=1}^{2022} f(\lambda+k)=\sum_{k=1}^{2022}(\lambda+k)$ $$ =\underbrace{\lambda+\lambda+\ldots+\lambda}_{2022}+(1+2+\ldots .+2022) $$ $=2022 \lambda+\frac{2022 	imes 2023}{2} \leq(2022)^{2}$ (Given)\ $\Rightarrow \lambda \leq \frac{2021}{2}$\ So, largest $\lambda=1010$\