Q.In the expansion of $(x + a)^n$ if the sum of odd terms is denoted by $O$ and the sum of even terms by $E$, then prove that

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 $y$? …