Animated Solution for Mathematics - Sequence and Series: The sum $1+2\cdot 3+3\cdot 3^2+\cdots +10\cdot 3^9$ is equal to

Mastering the AGP

My dear student, welcome to the arena. Today, we are not just solving a problem; we are dissecting a mathematical structure.

When you look at the series $1+2\cdot 3+3\cdot 3^2+\cdots +10\cdot 3^9$, what do you see? Do you see a chaotic mess of numbers, or do you see a hidden order?

In the world of JEE Advanced, the ability to see the order within chaos is what separates the aspirants from the achievers. Let us peel back the layers of this Arithmetic-Geometric Progression (AGP).

The Anatomy of the Series

First, let us identify our components. We have two sequences dancing together.

The first part is $1,2,3,\dots ,10$. This is an Arithmetic Progression (AP) with the first term $a=1$ and common difference $d=1$.

The second part is $1,3,3^2,\dots ,3^9$. This is a Geometric Progression (GP) with the first term $1$ and common ratio $r=3$.

When these two distinct species of sequences are multiplied term-by-term, they form an AGP. Recognizing this is your first victory. You have identified the enemy; now, let us formulate the strategy.

The Art of the Shift

We define our sum as:

This is our baseline. Now, we perform the 'Shift and Subtract' maneuver. We multiply the entire equation by the common ratio of the GP, which is $3$.

This gives us:

Notice how the powers of $3$ have shifted? This is intentional. We are aligning the terms so that when we subtract, the coefficients will simplify.

The Telescoping Collapse

Now, we subtract $3S$ from $S$. Imagine the terms lining up like soldiers.

We have $S-3S$ on the left. On the right, we subtract vertically: $1$ remains $1$. Then, $(2\cdot 3-1\cdot 3)$ becomes $1\cdot 3$, and $(3\cdot 3^2-2\cdot 3^2)$ becomes $1\cdot 3^2$.

This pattern continues beautifully until the second-to-last term. Because we shifted the series, the term $10\cdot 3^{10}$ is left hanging at the end of our subtracted equation.

So, our equation becomes:

The Final Simplification

Look at the bracketed expression: $(1+3+3^2+\cdots +3^9)$. This is a pure Geometric Progression with $10$ terms.

We use the standard sum formula for a GP:

Substituting our values, $a=1$, $r=3$, and $n=10$, we get:

Now, we substitute this back into our equation for $-2S$:

To combine these, we find a common denominator of $2$. This transforms the equation into:

Simplifying the numerator, we get:

Finally, multiply by $-1$ and divide by $2$ to isolate $S$. We arrive at the final result:

$$S=\frac{19\cdot 3^{10}+1}{4}$$

The Elegance of Method

Look at what we have achieved. We took a complex, intimidating series and, through the simple, elegant application of the 'Shift and Subtract' method, reduced it to a basic arithmetic calculation.

This is the essence of physics and mathematics in the JEE curriculum. It is rarely about brute force; it is about finding the right transformation. Keep this logic in your toolkit, and no series will ever intimidate you again. You have done well.