Animated Solution for Mathematics - Sequence and Series: If $(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+\cdots +10(11{)}^9=k(10{)}^9$, then $k$ is equal to:

Decoding the AGP

Welcome, future engineer. When you first look at the expression $(10{)}^9+2(11{)}^1(10{)}^8+3(11{)}^2(10{)}^7+\cdots +10(11{)}^9$, it is perfectly natural to feel a momentary spike of anxiety. It looks like a chaotic mess of powers and coefficients.

But in the world of JEE Advanced, chaos is often just order in disguise. Our job today is to peel back the layers of this expression and reveal the elegant structure underneath.

The Power of Factoring

The first step in any complex problem is to simplify the landscape. We are given the sum:

Notice that every term involves powers of $10$ and $11$, and the total power in each term is $9$. Let us factor out $(10{)}^9$ from the entire expression:

Suddenly, the problem breathes. Let us define $x=\frac{11}{10}=1.1$. Now, our inner series, which we will call $S^{\prime }$, looks like this:

This is the heartbeat of the problem.

Identifying the AGP

Look closely at $S^{\prime }$. We have coefficients $1,2,3,\dots ,10$, which form an Arithmetic Progression (AP) with a common difference of $1$.

Simultaneously, we have powers of $x$ ($1,x,x^2,\dots ,x^9$), which form a Geometric Progression (GP) with a common ratio of $x$. When an AP and a GP are multiplied term-by-term, we get an Arithmetic-Geometric Progression (AGP).

The Shift and Subtract Strategy

How do we sum an AGP? We use the 'Shift and Subtract' method. We write down $S^{\prime }$ and then write down $x{S}^{\prime }$, shifting every term one position to the right:

Now, subtract the second equation from the first. On the left, we get $S^{\prime }(1-x)$. On the right, the coefficients collapse:

The Final Simplification

The series $1+x+x^2+\cdots +x^9$ is a standard GP with $10$ terms. Its sum is $\frac{1-x^{10}}{1-x}$. Substituting this back, we get:

To isolate $S^{\prime }$, we divide by $(1-x)$:

Now, we bring back our value $x=1.1$, which implies $1-x=-0.1$. Substituting this:

Since $(-0.1{)}^2=0.01$, the first term becomes $100(1-1.1^{10})$. The second term, with the double negative, becomes $+100(1.1{)}^{10}$.

When we expand this, we get $100-100(1.1{)}^{10}+100(1.1{)}^{10}$. The terms involving $(1.1{)}^{10}$ cancel out perfectly, leaving $S^{\prime }=100$.

Since $S=S^{\prime }\times 10^9$, and the problem states $S=k(10{)}^9$, we conclude that $k=100$.

See how the complexity vanished? We didn't need a calculator or brute force. We needed structure, patience, and the courage to manipulate the algebra. You have just mastered the AGP. Keep this logic in your toolkit—it will serve you well in the exam hall.