Q.Find the sum of the infinite series $1 + \dfrac{1}{3} + \dfrac{1\cdot 3}{3\cdot 6} + \dfrac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9} + \cdots$

The Binomial Theorem: From Patterns to Power

Imagine you have to expand $(x + y)^2$. You know it's $x^2 + 2xy + y^2$. What about $(x + y)^3$? That's $x^3 + 3x^2y + 3xy^2 + y^3$. Now try $(x + y)^4$ — you could multiply $(x + y)^3$ by $(x + y)$ again, but it gets messy fast.

The Binomial Theorem is the shortcut. It tells you exactly what $(x + y)^n$ expands to, for any positive integer $n$, without doing the multiplication step by step.

The Pattern You Already Know

Look at the expansions we have:

Three things stand out:

1. The powers of $x$ decrease from $n$ down to $0$, while the powers of $y$ increase from $0$ up to $n$. In each term, the exponents add to $n$.
2. The coefficients — 1, 2, 1 for $n=2$; 1, 3, 3, 1 for $n=3$; 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

Why Do These Coefficients Appear?

Think about what $(x + y)^n$ really means. It's $(x + y)$ multiplied by itself $n$ times:

When you expand, you pick either $x$ or $y$ from each factor. A term like $x^{n-k}y^k$ comes from choosing $y$ from exactly $k$ of the $n$ factors and $x$ from the rest.

How many ways can you choose which $k$ factors give you $y$? That's exactly the number of combinations: $\binom{n}{k}$ (read "n choose k").

So the coefficient of $x^{n-k}y^k$ is $\binom{n}{k}$. That's the heart of the theorem.

The Precise Statement

Notice $\binom{n}{0} = 1$ and $\binom{n}{n} = 1$, which matches the first and last coefficients always being 1.

A Quick Example

Expand $(2a - b)^5$ using the theorem.

Here $x = 2a$, $y = -b$, and $n = 5$.

Compute term by term:

• $k=0$: $\binom{5}{0}(2a)^5(-b)^0 = 1 \cdot 32a^5 = 32a^5$
• $k=1$: $\binom{5}{1}(2a)^4(-b)^1 = 5 \cdot 16a^4 \cdot (-b) = -80a^4b$
• $k=2$: $\binom{5}{2}(2a)^3(-b)^2 = 10 \cdot 8a^3 \cdot b^2 = 80a^3b^2$
• $k=3$: $\binom{5}{3}(2a)^2(-b)^3 = 10 \cdot 4a^2 \cdot (-b^3) = -40a^2b^3$ …