Animated Solution for Mathematics - Sequence and Series: If $|x|<1,|y|<1$ and $x\ne y$, then the sum to infinity of the following series $(x+y)+(x^2+xy+y^2)+(x^3+x^{2}y+x{y}^2+y^3)+......is:$

Welcome, future engineers. Today, we are going to dismantle a problem that, at first glance, looks like a chaotic mess of variables. It is an infinite series, but not the kind you can solve by simply plugging in a formula.

It requires vision. It requires the ability to see the hidden structure beneath the surface. Imagine you are standing on the edge of this series:

It looks intimidating, doesn't it? But remember, in JEE Advanced, complexity is often just simplicity in disguise. Let's break this down.

The first step is to stop looking at the series as a whole and start looking at the individual terms. Let us define the $k$-th term as $T_k$.

The first term is $T_1=x+y$. The second is $T_2=x^2+xy+y^2$. The third is $T_3=x^3+x^{2}y+x{y}^2+y^3$.

Do you see the pattern? Each term is a homogeneous polynomial. There is a beautiful algebraic identity that governs these structures:

This identity is our golden key. It allows us to transform a sum of terms into a fraction. Because the problem guarantees that $x eq y$, we can use this identity without fear of dividing by zero.

Now that we have our tool, let's rewrite the general term $T_k$. Using our identity, we have:

Our total sum $S$ is simply the summation of these terms from $k=1$ to infinity:

Since $\frac{1}{x-y}$ is a constant relative to the summation index $k$, we can pull it out:

Now, we can distribute the summation:

We have successfully split one complex series into two distinct, manageable infinite geometric progressions.

Let's look at the first series: ${\sum }_{k=1}^{\infty }{x}^{k+1}=x^2+x^3+x^4+\dots$. This is a classic infinite GP with first term $a=x^2$ and common ratio $r=x$.

Since $|x|<1$, the sum is $\frac{x^2}{1-x}$. Similarly, the second series is ${\sum }_{k=1}^{\infty }{y}^{k+1}=y^2+y^3+y^4+\dots$, which sums to $\frac{y^2}{1-y}$.

Substituting these back, we get:

We are almost there. We just need to perform the algebraic cleanup. Finding the common denominator $(1-x)(1-y)$, the expression becomes:

Expanding the numerator gives $x^2-x^{2}y-y^2+x{y}^2$. Grouping terms, we get $(x^2-y^2)-(x^{2}y-x{y}^2)$, which factors into $(x-y)(x+y)-xy(x-y)$.

Factoring out $(x-y)$, we are left with $(x-y)(x+y-xy)$. Finally, we cancel the $(x-y)$ term from the numerator and denominator.

The result is:

This matches Option (B). By breaking the problem into smaller, logical steps, we turned a daunting infinite series into a simple algebraic expression. Keep practicing this mindset, and you will conquer any problem JEE throws at you.