Animated Solution for Mathematics - Binomial Theorem: If the expansion in powers of $x$ of the function $\frac{1}{(1-ax)(1-bx)}$ is $a_0+a_{1}x+a_2{x}^2+a_3{x}^3....$ then $a_n$ is

Unlocking the Power Series

Imagine you are standing before a complex, intimidating mathematical structure: the rational function $f(x)=\frac{1}{(1-ax)(1-bx)}$. It looks simple enough, yet it hides a secret.

We are tasked with finding the coefficient $a_n$ of the $x^n$ term in its infinite power series expansion. Many students, when faced with this, immediately reach for the binomial theorem, attempting to expand both $(1-ax{)}^{-1}$ and $(1-bx{)}^{-1}$ and then multiplying them.

While that path is technically correct, it is a treacherous journey through a forest of indices and summations. Today, we are going to take the elegant path—the path of the master.

The Scalpel of Algebra

Why struggle with the product in the denominator? The product $(1-ax)(1-bx)$ is the source of our headache.

In mathematics, when we see a product in the denominator, we should immediately think of Partial Fraction Decomposition. This is our scalpel. It allows us to slice through the complexity and separate the function into two independent, manageable pieces.

We assume that our function can be written as:

By multiplying both sides by the common denominator $(1-ax)(1-bx)$, we obtain the identity $1=A(1-bx)+B(1-ax)$. This is where the magic happens.

By choosing strategic values for $x$, we can isolate $A$ and $B$. If we set $x=\frac{1}{a}$, the second term vanishes, revealing $A=\frac{a}{a-b}$.

Similarly, setting $x=\frac{1}{b}$ gives us $B=\frac{-b}{a-b}$. We have successfully dismantled the beast!

The Heartbeat of Power Series

Now that we have $f(x)=\frac{1}{a-b}\left[\frac{a}{1-ax}-\frac{b}{1-bx}\right]$, the problem transforms. We are no longer dealing with a product; we are dealing with two standard geometric series.

Recall the fundamental identity:

This is the heartbeat of power series. For our first term, $\frac{a}{1-ax}$, we treat $ax$ as our $t$. The expansion becomes $a\sum (ax{)}^n$.

If we zoom in on the $n$-th term, we see $a\cdot (ax{)}^n$, which simplifies to $a^{n+1}{x}^n$. Do you see the beauty?

The coefficient is not just $a^n$; it is $a^{n+1}$ because of the $a$ we pulled out during decomposition. We apply the exact same logic to the second term, $\frac{b}{1-bx}$, yielding $b^{n+1}{x}^n$.

The Synthesis

We are at the finish line. We have the two components of our coefficient $a_n$. Combining them, we get:

This is the general expression for the $n$-th coefficient. To match this with our given options, we simply multiply the numerator and denominator by $-1$, giving us $\frac{b^{n+1}-a^{n+1}}{b-a}$.

This is the exact form we were looking for.

I know that at the start, this problem looked like a mountain of algebra. But by using partial fractions, we turned a multiplication problem into an addition problem, and by using the geometric series, we turned a rational function into a simple sequence.

This is the essence of JEE Advanced physics and math: it is not about brute force; it is about finding the right tool to reveal the underlying simplicity.

Keep practicing this mindset, and you will find that even the most terrifying problems have a beautiful, logical soul waiting to be discovered.