The Binomial Theorem – Complete Lesson

Introduction: Why Do We Need the Binomial Theorem?

My dear student, let me start by asking you a question. Have you ever tried to expand (x + y)⁵ by multiplying (x + y) by itself five times? If you tried, you know it takes a long time and it is very easy to make a mistake. Now imagine you need to expand (x + y)⁸₀ — that would be almost impossible to do by hand!

The Binomial Theorem gives us a beautiful shortcut. It tells us exactly what the expansion looks like without doing all that multiplication. It also tells us how to find any specific coefficient we want, such as the coefficient of \(x^5y^3\) in \((x + y)^9\), without expanding the whole thing.

This theorem is one of the most important results in combinatorics. It connects algebra with counting in a very deep way. Let us learn it step by step.

Part 1: Understanding Binomial Expansions

What is a Binomial?

A binomial is simply an algebraic expression with two terms. For example, \(x + y\), \(2a + 3b\), and \(x – 1\) are all binomials. When we raise a binomial to a power, we get a binomial expansion.

Before we state the theorem, let us look at small expansions and see if we can find a pattern:

Can You See the Pattern?

Look carefully at each expansion. I want you to notice three things:

First: The number of terms in each expansion is one more than the power. For example, \((x+y)^3\) has 4 terms, and \((x+y)^5\) has 6 terms.

Second: In each term, the sum of the powers of \(x\) and \(y\) equals the original power. For example, in \((x+y)^4\), every term has powers that add up to 4: \(x^4y^0\), \(x^3y^1\), \(x^2y^2\), \(x^1y^3\), \(x^0y^4\).

Third: The coefficients follow a special pattern. Let me write just the coefficients:

Do you recognize this triangle? This is the famous Pascal’s Triangle! Each number is the sum of the two numbers directly above it. For example, in row 4: 6 = 3 + 3, and 4 = 1 + 3.

Now, these coefficients have a mathematical name. They are called binomial coefficients, and they are written as \(\binom{n}{r}\), which we read as “n choose r“.

Part 2: The Binomial Theorem — Formal Statement

The Theorem

Let me explain each part clearly:

• \(\binom{n}{r}\) — the binomial coefficient, equals \(\frac{n!}{r!(n-r)!}\)
• \(x^{n-r}\) — the power of \(x\) starts at \(n\) and decreases by 1 each term
• \(y^r\) — the power of \(y\) starts at 0 and increases by 1 each term
• The sum of powers in every term is always \((n-r) + r = n\)
• The number of terms is \(n + 1\) (from \(r = 0\) to \(r = n\))

Example 1: Expand \((x + y)^4\)

Solution: Using the Binomial Theorem with \(n = 4\):

Now calculate each binomial coefficient:

• \(\binom{4}{0} = 1\)
• \(\binom{4}{1} = 4\)
• \(\binom{4}{2} = \frac{4!}{2! \cdot 2!} = \frac{24}{4} = 6\)
• \(\binom{4}{3} = 4\) (by symmetry: \(\binom{4}{3} = \binom{4}{1}\))
• \(\binom{4}{4} = 1\)

So the expansion is:

This matches what we got by direct multiplication! But the theorem gives us the answer directly without multiplying.

Example 2: Expand \((x + y)^6\)

Solution: Using the theorem with \(n = 6\):

Calculate the coefficients:

• \(\binom{6}{0} = 1\)
• \(\binom{6}{1} = 6\)
• \(\binom{6}{2} = \frac{6!}{2! \cdot 4!} = 15\)
• \(\binom{6}{3} = \frac{6!}{3! \cdot 3!} = \frac{720}{36} = 20\)
• \(\binom{6}{4} = \binom{6}{2} = 15\) (symmetry)
• \(\binom{6}{5} = \binom{6}{1} = 6\) (symmetry)
• \(\binom{6}{6} = 1\)

Part 3: Finding Specific Coefficients

One of the most common exam questions is: “Find the coefficient of a particular term in a binomial expansion.” You do NOT need to expand the whole thing. Let me show you the technique.

Example 3: Coefficient of \(x^5y^4\) in \((x + y)^9\)

Solution: We want the term where the power of \(x\) is 5 and the power of \(y\) is 4.

From the general term \(\binom{n}{r}x^{n-r}y^r\), we need:

• \(n – r = 5\) (power of \(x\))
• \(r = 4\) (power of \(y\))
• \(n = 9\) (given)

Check: \(n – r = 9 – 4 = 5\). ✓ This is consistent!

The coefficient is simply \(\binom{9}{4}\):

So the coefficient of \(x^5y^4\) in \((x+y)^9\) is 126.

Example 4: Coefficient of \(x^3y^2\) in \((2x – 3y)^5\)

Solution: This one is trickier because there are coefficients (2 and −3) inside the binomial. Let me handle it carefully.

Write \((2x – 3y)^5\) as \((2x + (-3y))^5\). Now apply the theorem:

We want the term where the power of \(x\) is 3 and the power of \(y\) is 2:

• Power of \(x\): \(5 – r = 3\), so \(r = 2\)
• Power of \(y\): \(r = 2\) ✓

The term when \(r = 2\) is:

The coefficient is 720.

Notice: We must include the coefficients from inside the binomial (2 and −3). Many students forget this and only calculate \(\binom{5}{2} = 10\), which would be wrong!

Example 5: Coefficient of \(x^7\) in \((1 + x)^{11}\)

Solution: This is a special case where one variable is 1. The expansion is:

We want the term with \(x^7\), so \(r = 7\). The coefficient is:

The coefficient of \(x^7\) is 330.

Why did I use \(\binom{11}{4}\) instead of \(\binom{11}{7}\)? Because of the symmetry property \(\binom{n}{r} = \binom{n}{n-r}\). It is easier to compute \(\binom{11}{4}\) since the numbers are smaller: \(\frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330\).

Exam Questions on Finding Coefficients

Part 4: Important Corollaries of the Binomial Theorem

Corollary 1: Sum of All Binomial Coefficients

What happens if we substitute \(x = 1\) and \(y = 1\) in the Binomial Theorem?

Why is this important? It tells us, for example, that if we add all the coefficients in \((x+y)^{10}\), we get \(2^{10} = 1024\). This also has a combinatorial meaning: the total number of subsets of a set with \(n\) elements is \(2^n\).

Corollary 2: Alternating Sum of Binomial Coefficients

What if we substitute \(x = 1\) and \(y = -1\)?

This means the sum of coefficients in even positions equals the sum of coefficients in odd positions!

Corollary 3: Sum of Even-Position and Odd-Position Coefficients

From the two results above, we can derive:

Each sum equals \(2^{n-1}\) (half of \(2^n\)) when \(n \geq 1\).

Part 5: Pascal’s Identity

The Identity

This identity is exactly the rule for building Pascal’s Triangle! Each entry is the sum of the two entries above it. For example, \(\binom{5}{2} = \binom{4}{1} + \binom{4}{2} = 4 + 6 = 10\).

Algebraic Proof

Let me prove this identity algebraically. Starting from the right side:

Get a common denominator of \(r!(n-r)!\):

✓ This proves the identity.

Part 6: The Multinomial Theorem

When We Have More Than Two Terms

The Binomial Theorem works for expressions with two terms. What if we have three or more terms, like \((x + y + z)^4\)? For this, we need the Multinomial Theorem.

Example 6: Expand \((x + y + z)^3\)

Solution: We need all non-negative integer solutions to \(n_1 + n_2 + n_3 = 3\):

Example 7: Coefficient in a Multinomial Expansion

Problem: What is the coefficient of \(u^2w^3x^4y^2\) in \((u + v + 2w + x + 3y + z)^{11}\)?

Solution: The general term is:

We need: \(n_1 = 2\) (for \(u^2\)), \(n_3 = 3\) (for \(w^3\)), \(n_4 = 4\) (for \(x^4\)), \(n_5 = 2\) (for \(y^2\)), \(n_2 = 0\) (no \(v\)), \(n_6 = 0\) (no \(z\)).

Check: \(2 + 0 + 3 + 4 + 2 + 0 = 11\) ✓

The coefficient is:

Part 7: Special Values and Useful Identities

Some Special Substitutions

By substituting specific values for \(x\) and \(y\) in the Binomial Theorem, we can derive many useful identities:

Example 8: Evaluating a Sum

Problem: Find the value of \(\sum_{r=0}^{n} \binom{n}{r} 3^r\).

Solution: Compare with the Binomial Theorem: \((x + y)^n = \sum \binom{n}{r} x^{n-r} y^r\).

Set \(x = 1\) and \(y = 3\):

So the answer is \(4^n\).