If a function f satisfies f(m+n) = f(m) + f(n | vedclass

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(A) Given the functional equation $f(m+n) = f(m) + f(n)$ for all $m, n \in \mathbb{N}$.This is Cauchy's functional equation on natural numbers,which i

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$1010$

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(A) Given the functional equation $f(m+n) = f(m) + f(n)$ for all $m, n \in \mathbb{N}$.This is Cauchy's functional equation on natural numbers,which implies $f(x) = cx$.Given $f(1) = 1$,we have $c(1) = 1$,so $c = 1$.Thus,$f(x) = x$.Now,we need to solve the inequality $\sum_{k=1}^{2022} f(\lambda+k) \leq (2022)^2$.Substituting $f(x) = x$,we get $\sum_{k=1}^{2022} (\lambda+k) \leq (2022)^2$.Expanding the summation: $2022\lambda + \sum_{k=1}^{2022} k \leq (2022)^2$.Using the sum formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$,we have $2022\lambda + \frac{2022 	imes 2023}{2} \leq (2022)^2$.Dividing by $2022$: $\lambda + \frac{2023}{2} \leq 2022$.$\lambda \leq 2022 - 1011.5$.$\lambda \leq 1010.5$.Since $\lambda$ must be a natural number,the largest value is $\lambda = 1010$.

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 ..........

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