Animated Solution for Mathematics - Sequence and Series: $(\frac{1}{3}+\frac{4}{7})+(\frac{1}{3^2}+\frac{1}{3}\times \frac{4}{7}+\frac{4^2}{7^2})+(\frac{1}{3^3}+\frac{1}{3^2}\times \frac{4}{7}+\frac{1}{3}\times \frac{4^2}{7^2}+\frac{4^3}{7^3})+\dots$ upto infinite terms, is equal to

Welcome, fellow traveler on the road to JEE Advanced. Today, we are not just solving a problem; we are uncovering a hidden symmetry.

When you first look at the expression $(\frac{1}{3}+\frac{4}{7})+(\frac{1}{3^2}+\frac{1}{3}\times \frac{4}{7}+\frac{4^2}{7^2})+\dots$, it might look like a chaotic mess of fractions. But in mathematics, chaos is often just order waiting to be discovered.

Imagine you are staring at a complex machine. If you try to fix every gear at once, you will get overwhelmed. Instead, let us simplify.

Let $a=\frac{1}{3}$ and $b=\frac{4}{7}$. Suddenly, the series transforms into something elegant:

Do you see it now? Each bracket is a finite geometric progression. This is the beauty of abstraction—it clears the fog so we can see the underlying structure.

We need a way to condense these brackets. Recall the classic identity for the difference of powers:

If we rearrange this, we find that the sum of the terms inside our $n$-th bracket is simply:

By introducing this identity, we have turned a multi-term summation into a clean, singular expression. We are no longer dealing with a list of fractions; we are dealing with a function of $n$.

Now, we calculate the constant factor outside our expression. With $b-a=\frac{4}{7}-\frac{1}{3}=\frac{12-7}{21}=\frac{5}{21}$, we know that $\frac{1}{b-a}=\frac{21}{5}$.

Our total sum $S$ is now:

This is where the magic happens. We can split this into two separate infinite geometric series:

For the first series, $S_b=b^2+b^3+b^4+\dots$, the first term is $b^2$ and the ratio is $b$. Using the infinite sum formula $S=\frac{\text{first term}}{1-\text{ratio}}$, we get:

Similarly, for $S_a=\frac{a^2}{1-a}$, substituting $a=\frac{1}{3}$ yields:

Finally, we bring it all home.

Finding the common denominator of $42$, we get:

The terms cancel out with breathtaking precision: $\frac{21}{42}$ becomes $\frac{1}{2}$, and $\frac{25}{5}$ becomes $5$. Our final answer is $\frac{5}{2}$.