Animated Solution for Mathematics - Sequence and Series: $\frac{6}{3^{26}}+\frac{10\cdot 1}{3^{25}}+\frac{10\cdot 2}{3^{24}}+\frac{10\cdot 2^2}{3^{23}}+\cdots +\frac{10\cdot 2^{24}}{3}$ is equal to :

Welcome, fellow traveler on the path to JEE mastery. Today, we are going to dismantle a series that, at first glance, looks like a chaotic mess of powers and fractions.

Many students see a problem like $S=\frac{6}{3^{26}}+\frac{10\cdot 1}{3^{25}}+\frac{10\cdot 2}{3^{24}}+\cdots +\frac{10\cdot 2^{24}}{3}$ and immediately feel the urge to panic. But I want you to take a deep breath. In the world of competitive mathematics, complexity is often just a mask for elegance.

Let us peel back that mask together.

The Outlier

The first step in any series problem is to identify the pattern. If you look closely, the first term, $\frac{6}{3^{26}}$, stands out. It does not share the $10$ in the numerator that the rest of the terms possess.

This is our first clue. In mathematics, when a term breaks the symmetry, it is often an 'outlier' that needs to be handled separately. So, let us isolate it:

By separating the series, we have cleared the fog. Now, we can focus on the bracketed part, which is where the real beauty lies.

The General Term

To tame this series, we need to express it in the language of sigma notation. Let us find the $r^{th}$ term, $T_r$.

The numerator is always $10$, and the power of $2$ follows the sequence $2^0,2^1,2^2,\dots ,2^{24}$. This means the $r^{th}$ term has $2^{r-1}$ in the numerator. The denominator is a power of $3$ that decreases from $3^{25}$ down to $3^1$, which we can write as $3^{26-r}$.

Thus, our sum becomes:

The Exponent Dance

Now, we need to simplify the expression inside the sigma. We have a constant $10$ and a denominator $3^{26-r}$.

Let us pull the constant $10$ out and rewrite the denominator using exponent rules: $3^{26-r}=3^{26}\cdot 3^{-r}$. When we move $3^{-r}$ to the numerator, it becomes $3^r$. Our expression now looks like this:

This is where the magic happens. We have $2^{r-1}$ and $3^r$. To combine these into a single base, we need the exponents to be the same. We can borrow one factor of $3$ from $3^r$ to write it as $3\cdot 3^{r-1}$.

Now, we have:

The GP Revelation

Look at that summation: ${\sum }_{r=1}^{25}{6}^{r-1}$. This is a textbook Geometric Progression (GP) where the first term $a=1$, the common ratio $r=6$, and the number of terms $n=25$.

Using the sum formula $S_n=\frac{a(r^n-1)}{r-1}$, we get:

The Grand Finale

We are almost there. Let us substitute this back into our equation for $S$:

Since $30/5=6$, the expression simplifies beautifully to:

The $6$ and $-6$ cancel out, leaving us with $\frac{6^{26}}{3^{26}}$. This is simply $(\frac{6}{3}{)}^{26}$, which results in:

And there you have it! What seemed like a terrifying series collapsed into an elegant power of two. This is the power of persistence and pattern recognition. Keep practicing, and soon, you will see the beauty in every equation.