Q.Evaluate $\left(\sqrt{3} + \sqrt{2}\right)^6 - \left(\sqrt{3} - \sqrt{2}\right)^6$.

The Problem That Started It All

Imagine you have to expand $(x + y)^2$. That's easy: $x^2 + 2xy + y^2$. Now try $(x + y)^3$: $x^3 + 3x^2y + 3xy^2 + y^3$. Still manageable.

But what about $(x + y)^{10}$? Or $(x + y)^{100}$? Multiplying it out term by term would take forever. There has to be a pattern — and there is.

The Binomial Theorem is the shortcut that tells you exactly what each term in the expansion of $(x + y)^n$ looks like, without ever having to multiply.

The Pattern You Already Know

Look at the expansions you already know:

Notice three things:

1. The powers of $x$ decrease from $n$ down to $0$, while the powers of $y$ increase from $0$ up to $n$. In every term, the exponents add to $n$.
2. The coefficients — 1, 4, 6, 4, 1 for $n=4$ — follow a famous pattern called Pascal's triangle.
3. The number of terms is always $n+1$.

       1
      1 1
     1 2 1
    1 3 3 1
   1 4 6 4 1
  1 5 10 10 5 1

The Precise Statement

Let's break that down.

The symbol $\sum_{k=0}^{n}$ means "add up terms for $k = 0, 1, 2, \dots, n$". For each $k$, the term is:

• $\binom{n}{k}$ — the coefficient (read as "n choose k")
• $x^{n-k}$ — $x$ raised to the power $n-k$
• $y^k$ — $y$ raised to the power $k$

So for $n=4$, the terms are:

Add them up: $x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$. Exactly what we had.

Where Do Those Coefficients Come From?

The binomial coefficient $\binom{n}{k}$ counts how many ways you can choose $k$ items from a set of $n$ items. In the expansion, it counts how many ways you can pick $k$ copies of $y$ (and therefore $n-k$ copies of $x$) when multiplying $(x+y)$ by itself $n$ times.

The General Term

The $k$-th term (starting with $k=0$) in the expansion is:

This is the most useful part for exams. If someone asks "find the 5th term in $(x+y)^{10}$", you set $k=4$ (because $T_{k+1}$ means $k=4$ gives the 5th term) and write:

What If It's Not Just $x$ and $y$?

The theorem works for any expression. For $(2a - 3b)^5$, treat $x = 2a$ and $y = -3b$:

The $k$-th term becomes $\binom{5}{k} (2)^{5-k} (-3)^k a^{5-k} b^k$. The coefficients get multiplied by powers of 2 and -3.

Why This Matters

The Binomial Theorem isn't just a formula — it's a window into combinatorics, probability, and even calculus. It lets you:

• Expand any binomial instantly
• Find a specific term without expanding everything
• Approximate values like $(1.01)^{10}$ by setting $x=1$, $y=0.01$
• Understand the binomial distribution in statistics

The core idea: every term in $(x+y)^n$ has the form $\binom{n}{k} x^{n-k} y^k$, and the theorem tells you exactly which $k$ to use and what coefficient goes with it.

Expanding binomial expressions using the Binomial Theorem is a core topic in the NCERT Class 11 Mathematics chapter on Binomial Theorem, and "binomial theorem expansion formula and examples" is one of the most searched topics for CBSE board and JEE Main preparation. Finding a specific general term without full expansion is also a classic question type that appears repeatedly in "binomial theorem important questions" for competitive exams.