Animated Solution for Mathematics - Sequence and Series: Let $S=N\cup \{0\}$. Define a relation R from S to R by: $R=\{(x,y):{\log }_{e}y=x{\log }_e(\frac{2}{5}),x\in S,y\in R\}$. Then, the sum of all the elements in the range of R is equal to

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dissect a problem that might look like a standard algebraic exercise, but beneath the surface, it is a beautiful lesson in how we bridge the gap between abstract logarithmic definitions and concrete arithmetic series.

Often, students see a logarithm and immediately panic, looking for complex identities. But in the world of JEE Advanced, the most powerful tool is often the simplest: clarity of thought.

Let us begin by looking at the set $S=\mathbb{N}\cup \{0\}$. In the language of mathematics, $\mathbb{N}$ represents the natural numbers—the counting numbers $1,2,3,\dots$. By taking the union with the set containing zero, we have defined $S$ as the set of whole numbers: $S=\{0,1,2,3,\dots \}$.

Why is this important? Because the relation $R$ is defined by the variable $x$ belonging to this set. If $S$ were the set of all real numbers, we would be dealing with a continuous curve.

But because $S$ is discrete, our relation $R$ is not a line; it is a collection of isolated points. When you see a domain like this, stop and visualize it. You are not drawing a graph; you are generating a sequence.

Now, let us confront the relation itself: ${\log }_{e}y=x{\log }_e(\frac{2}{5})$. This equation is the heart of the problem. It connects our input $x$ to our output $y$.

Many students see the logarithm and feel the urge to start calculating values immediately. Resist that urge! Instead, look for the structure. We have a coefficient $x$ multiplying a logarithm, which invokes the power rule of logarithms: $n\log a=\log a^n$.

By applying this rule, we transform the right side of our equation into ${\log }_e(\frac{2}{5}{)}^x$. Now, our equation looks like this:

Because the logarithmic function is one-to-one, we can confidently equate the arguments. The logs vanish, leaving us with a clean, elegant exponential function:

This is the "Aha!" moment. We have successfully stripped away the complexity of the logarithm to reveal a simple exponential relationship. We are no longer doing log problems; we are doing sequence problems.

With our formula $y=(\frac{2}{5}{)}^x$ in hand, we can now generate the elements of the range. We simply plug in the values of $x$ from our set $S$:

For $x=0$, $y=(\frac{2}{5}{)}^0=1$. For $x=1$, $y=(\frac{2}{5}{)}^1=\frac{2}{5}$. For $x=2$, $y=(\frac{2}{5}{)}^2=\frac{4}{25}$.

As we continue this process for $x=3,4,\dots$, we generate the sequence: $1,\frac{2}{5},\frac{4}{25},\frac{8}{125},\dots$. This is a geometric progression where each term is obtained by multiplying the previous term by a common ratio $r=\frac{2}{5}$.

The problem asks for the sum of all elements in the range. Since our domain $S$ is infinite, we are looking for the sum of an infinite geometric series:

We know that for an infinite geometric progression with first term $a$ and common ratio $r$, the sum $S_{\infty }$ is given by the formula:

Here, our first term $a=1$, and our common ratio $r=\frac{2}{5}$. Since $|\frac{2}{5}|<1$, the series converges beautifully. Let us perform the final calculation:

Dividing by a fraction is equivalent to multiplying by its reciprocal. Thus, our final result is:

Look at what we have achieved. We started with a logarithmic relation that seemed daunting, identified the discrete nature of the domain, transformed the equation into a simple exponential form, and recognized the underlying geometric progression.

This is the essence of JEE Advanced mathematics: it is not about memorizing formulas, but about recognizing patterns and simplifying the complex into the manageable. You have successfully navigated the logic, and the result, $\frac{5}{3}$, is your reward.