Animated Solution for Mathematics - Sequence and Series: Let $f:\mathbb{R}\to \mathbb{R}$ be a function which satisfies $f(x+y)=f(x)+f(y)\forall x,y\in \mathbb{R}$. If $f(1)=2$ and $g(n)={\sum }_{k=1}^{n-1}f(k)$, $n\in \mathbb{N}$ then the value of $n$, for which $g(n)=20$, is :

Welcome, student. Today, we are not just solving a problem; we are peeling back the layers of a mathematical structure that appears frequently in the JEE Advanced arena. When you first look at the equation $f(x+y)=f(x)+f(y)$, you might feel a slight hesitation.

It looks abstract, doesn't it? But this is not a random collection of symbols. This is the legendary Cauchy's Functional Equation.

Think of this equation as a definition of linearity. It tells us that the function respects the additive structure of its input. If you add two inputs, the function adds their outputs.

In the world of continuous functions, there is only one family of functions that behaves this way: the linear family, $f(x)=cx$. Imagine a straight line passing through the origin. If you take any two points on that line, the sum of the function values at those points will always equal the function value of the sum of the points. It is a beautiful, rigid, and predictable property.

Now that we have identified the soul of the function as $f(x)=cx$, our next mission is to find the constant $c$. We are given the condition $f(1)=2$. This is our anchor.

By substituting $x=1$ into our general form, we get $c(1)=2$, which immediately reveals that $c=2$.

So, our function is not just any linear function; it is the specific function $f(x)=2x$. Every time you see an input, you simply double it. It is elegant, it is simple, and it is the key to everything that follows.

Now, let us turn our attention to the second part of the problem: $g(n)={\sum }_{k=1}^{n-1}f(k)$. This looks like a daunting summation, but remember, we have already decoded $f(k)$. Since $f(x)=2x$, it follows that $f(k)=2k$.

Substituting this into our summation, we get:

We can pull the constant $2$ out of the summation, which is a standard property of sigma notation. This leaves us with:

Now, recall the classic formula for the sum of the first $m$ natural numbers:

In our specific case, the upper limit of the summation is not $n$, but $n-1$. So, we set $m=n-1$. Substituting this into the formula, we get:

Watch the magic happen here. The $(n-1)+1$ simplifies beautifully to $n$. The $2$ in the numerator cancels perfectly with the $2$ in the denominator. We are left with a clean, quadratic expression:

We are almost at the finish line. The problem tells us that $g(n)=20$. So, we set our expression equal to $20$:

Expanding this, we get $n^2-n=20$, or more formally, $n^2-n-20=0$. This is a standard quadratic equation. We need to find two numbers that multiply to $-20$ and add up to $-1$. Those numbers are $-5$ and $+4$.

Thus, we factor the equation as:

This gives us two potential solutions: $n=5$ and $n=-4$.

In the heat of an exam, it is easy to stop here. But a true JEE aspirant always checks the constraints. The problem explicitly states that $n\in \mathbb{N}$, meaning $n$ must be a natural number.

Natural numbers are positive integers. Therefore, $n=-4$ is physically impossible in this context. We reject it without a second thought.

Our final answer is $n=5$.