Animated Solution for Mathematics - Sequence and Series: The sum to infinite term of the series $1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots$ is

Unlocking the AGP

Welcome, future engineer. Today, we are going to peel back the layers of a series that, at first glance, might seem like a chaotic jumble of numbers. We are looking at the series $S=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+\dots$.

When you see a problem like this in the JEE Advanced, your first instinct might be to panic—how do we sum an infinite number of terms that don't follow a simple, single pattern? The secret, my friend, is to stop looking at the series as a whole and start looking at its DNA.

The Detective Work

Let's play detective. Look at the numerators: $1,2,6,10,14,\dots$. If we ignore that first $1$ for a moment, the sequence $2,6,10,14$ is clearly an Arithmetic Progression (AP) with a common difference of $d=4$.

Now, look at the denominators: $3^1,3^2,3^3,3^4,\dots$. This is a classic Geometric Progression (GP) with a common ratio of $r=\frac{1}{3}$.

When you have an AP multiplied by a GP, you have an Arithmetico-Geometric Progression (AGP). This is a beautiful structure because it contains the growth of an AP and the decay of a GP. The key to solving it is not to fight the complexity, but to use the GP's common ratio to 'shift' the series and force it to collapse.

The Shift and Subtract Strategy

Imagine the series $S$ as a long, infinite train. To simplify it, we are going to create a second train, $\frac{1}{3}S$, which is just the original train shifted by one carriage.

We write:

Now, multiply every term by $\frac{1}{3}$ and shift it to the right:

Why do we do this? Because when we subtract the second equation from the first, the denominators align perfectly. We are essentially subtracting the 'growth' of the AP from itself. Let's perform the subtraction $S-\frac{1}{3}S$:

Look at the result. The numerators are no longer $2,6,10,14$. They have become $1,4,4,4,\dots$. The subtraction has effectively 'killed' the AP part, leaving us with a constant difference!

The Collapse

We are left with:

Combine the first two terms: $1+\frac{1}{3}=\frac{4}{3}$. Now, look at the remaining infinite tail: $\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+\dots$. We can factor out the $4$:

Inside that bracket is a pure, infinite Geometric Progression. The first term $a=\frac{1}{3^2}=\frac{1}{9}$, and the common ratio $r=\frac{1}{3}$. We know the sum of an infinite GP is $S_{\infty }=\frac{a}{1-r}$.

Let's calculate that sum:

The Final Victory

We are almost there. Substitute this value back into our equation:

Finally, solve for $S$:

And there it is. The infinite sum is exactly $3$.

Think about what we just did. We took an infinite, complex series and, through the simple act of shifting and subtracting, forced it to reveal its true, finite nature. This is the essence of mathematics—finding order in apparent chaos. Keep this 'shift and subtract' technique in your toolkit; it is the key to unlocking almost any AGP you will encounter in your journey. You've got this!