If a function f satisfies f(m + n) = f(m) + f(n) for all m, n ∈ N and f(1) = 1, then the largest nat

Q: Wait, how do we know $f(n) = n$ for sure? Could it be $f(n) = cn$?

Excellent observation! For the domain of natural numbers $\mathbb{N}$, the Cauchy equation $f(m+n) = f(m) + f(n)$ implies $f(n) = cn$. However, the problem explicitly gives us the initial condition $f(1) = 1$. Substituting $n=1$ gives $f(1) = c(1) = 1$, which forces $c=1$. Thus, $f(n) = n$ is the only solution.

Q: Why can we just split the summation $\sum_{k=1}^{2022} (\lambda + k)$ into two parts?

This is due to the linearity property of the summation operator. The sum of a sum is the sum of the sums. Specifically, $\sum (a_k + b_k) = \sum a_k + \sum b_k$. Here, $\lambda$ is constant with respect to $k$, so it behaves like a scalar multiplier in the summation.

Q: I got $\lambda \le 1010.5$. Why is the answer 1010 and not 1010.5?

The problem asks for the largest *natural number* $\lambda$. Natural numbers are integers $\{1, 2, 3, \dots\}$. Since $1010.5$ is not a natural number, we must find the largest integer that satisfies the inequality $\lambda \le 1010.5$. That integer is $1010$.

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 $\sum_{k=1}^{2022} f(\lambda + k) \le (2022)^2$ is equal to _________

Sigma Percentile · Accepted Answer

### The Beauty of Functional Equations

Welcome, fellow traveler on the road to JEE Advanced. Today, we are going to dissect a problem that, at first glance, might look like a daunting algebraic puzzle.

We are presented with a functional equation: $f(m + n) = f(m) + f(n)$. This is not just any equation; this is the legendary **Cauchy Functional Equation**.

It is the bedrock of linear behavior in mathematics. When you see this, I want you to feel a sense of calm. It is telling you that the function is additive—the whole is exactly the sum of its parts.

### Decoding the Function

Let us start by peeling back the layers. We are given $f(1) = 1$.

If we test $m=1$ and $n=1$, we get $f(1+1) = f(1) + f(1)$, which implies $f(2) = 1 + 1 = 2$. If we continue this, testing $f(3) = f(2+1) = f(2) + f(1) = 2 + 1 = 3$, the pattern becomes undeniable.

By the principle of mathematical induction, we can confidently state that $f(n) = n$ for all $n \in \mathbb{N}$. The function is simply the identity function! This realization is our first major victory.

### The Summation Challenge

Now, the problem shifts gears. We are asked to find the largest natural number $\lambda$ such that:

$$\sum_{k=1}^{2022} f(\lambda + k) \le (2022)^2$$

Since we have already established that $f(x) = x$, we can substitute this directly into our inequality. The expression becomes:

$$\sum_{k=1}^{2022} (\lambda + k) \le (2022)^2$$

This looks like a lot of terms, but do not panic. We use the linearity of the summation operator. We can split this into two distinct sums:

$$\sum_{k=1}^{2022} \lambda + \sum_{k=1}^{2022} k \le (2022)^2$$

### The Algebraic Shortcut

Here is where the magic happens. Many students will immediately try to calculate the sum of the first $2022$ integers. While that is correct, it is inefficient.

Let us look at the first term: $\sum_{k=1}^{2022} \lambda$. Since $\lambda$ does not depend on $k$, we are simply adding $\lambda$ to itself $2022$ times. This is just $2022\lambda$.

For the second term, we use the standard formula for the sum of the first $n$ natural numbers, which is $\frac{n(n+1)}{2}$. With $n = 2022$, we have:

$$\sum_{k=1}^{2022} k = \frac{2022 	imes 2023}{2}$$

Substituting these back into our inequality, we get:

$$2022\lambda + \frac{2022 	imes 2023}{2} \le (2022)^2$$

### The Final Simplification

Look at the equation again. Do you see the common factor of $2022$ in every single term? This is a gift.

We can divide the entire inequality by $2022$ to simplify our lives significantly:

$$\lambda + \frac{2023}{2} \le 2022$$

Now, the arithmetic becomes trivial. We know that $\frac{2023}{2} = 1011.5$. So our inequality is:

$$\lambda + 1011.5 \le 2022$$

Subtracting $1011.5$ from both sides, we arrive at:

$$\lambda \le 1010.5$$

### The Conclusion

We are almost there. The problem asks for the largest **natural number** $\lambda$.

Since $\lambda$ must be an integer and $\lambda \le 1010.5$, the largest possible value is clearly **$1010$**.

Take a moment to appreciate what we just did. We took a functional equation, reduced it to a simple identity, transformed a complex summation into a linear inequality, and used algebraic properties to avoid tedious calculations. This is the essence of JEE Advanced problem-solving.

Animated Solution for Mathematics - Functions: If a function $f$ satisfies $f (m + n) = f (m) + f (n)$ for all $m, n \in N$ and $f (1) = 1$ , then the largest natural number $λ$ such that $\sum_{k = 1}^{2022} f (λ + k) \leq (2022)^{2}$ is equal to _________

Enter Numerical Value:

Visualized Solution

The Functional Equation

Applying the Initial Condition

Generalizing $f (n)$

The Summation Inequality

Splitting the Summation

Evaluating the Constant Term

Sum of First $n$ Natural Numbers

Reconstructing the Inequality

Simplifying the Inequality

Solving for $λ$

The Final Answer

The Sigma Insight: Classification of Functions

The Beauty of Functional Equations

Decoding the Function

The Summation Challenge

The Algebraic Shortcut

The Final Simplification

The Conclusion

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