Skip to lesson content

Lesson 1 of 3

Binomial Theorem · Lesson 1 of 3

Discovering the Binomial Theorem

“See how coefficient patterns become a general expansion rule, and understand why that rule works.”

Learning Objectives

• Recognise the coefficient and exponent patterns in powers of a binomial. • Construct Pascal’s triangle and express its entries using combinations. • Use the binomial theorem and interpret its summation notation. • Follow the proof by mathematical induction and explain how adjacent coefficients combine. • Derive expansions for a difference and for 1 plus or minus a variable. • Obtain ordinary and alternating sums of binomial coefficients.

A familiar pattern with a bigger purpose

You already know that (a + b)² = a² + 2ab + b². Multiplying out a cube is also manageable, but a fifth or twelfth power would require many repeated products. The binomial theorem gives a systematic way to organise them. Before learning its formula, we will discover the pattern that the formula records.

Definition
Binomial and index

A binomial is an expression written as the sum or difference of two terms, such as a + b, x − 2y, or x² + 3/x. In (a + b)ⁿ, n is the exponent, also called the index. This chapter develops the theorem for positive integer n.

For a numerical calculation, choosing a convenient binomial can make powers easier to handle: 98² = (100 − 2)² and 999³ = (1000 − 1)³. The same idea is useful for higher powers such as 98⁵ and 101⁶. Our goal is an exact finite expansion for positive integer powers. We will not extend the formula to fractional or negative indices here.

IndexExpansionCoefficient row
0(a + b)⁰ = 1, when a + b ≠ 01
1a + b1, 1
2a² + 2ab + b²1, 2, 1
3a³ + 3a²b + 3ab² + b³1, 3, 3, 1
4a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴1, 4, 6, 4, 1

Look separately at the coefficient row and the powers attached to it. For a power n, the standard expansion has n + 1 positions. The power of a starts at n and decreases by one at each position. The power of b starts at zero and increases by one. In every position the two exponents add to n. This describes the formal expansion; after substituting particular expressions, some terms may vanish or combine.

Example — Reading the fourth-power pattern

Problem
Explain the structure of (a + b)⁴ without yet using a general formula.

  1. 1.There are five positions, because the power is 4.
  2. 2.The pairs of exponents of a and b are (4, 0), (3, 1), (2, 2), (1, 3), and (0, 4). Each pair sums to 4.
  3. 3.Attach the coefficient row 1, 4, 6, 4, 1 to those positions.
  4. 4.The result is a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴. A term such as 6a²b³ would be wrong because its exponents sum to 5.

Building Pascal’s triangle

The coefficient rows do more than list answers. They tell us how the next row can be created from the previous one. Put 1 at each end, then obtain every interior entry by adding the two adjacent entries immediately above it. This arrangement of rows is called Pascal’s triangle.

Coefficient rows of Pascal’s trianglen = 01n = 111n = 2121n = 31331n = 414641n = 515101051
Coefficient rows of Pascal’s triangle— Each row is labelled by its exponent n. The entries run from r = 0 to r = n. Compare the numerical entries with the combination notation.

For example, the row 1, 3, 3, 1 gives the next row 1, 4, 6, 4, 1 because the interior sums are 1 + 3, 3 + 3, and 3 + 1. Repeating the process gives 1, 5, 10, 10, 5, 1. The rows are labelled from n = 0, so the row for power 5 is the sixth displayed row. Distinguish the exponent label from the physical row count.

Two contributions make the next coefficientRow 4Row 546104 + 6 = 10
How adjacent entries generate the next row— The interior entry 10 is obtained by adding the adjacent 4 and 6 above it. The same rule produces every interior entry.

Why does adjacent addition work? When an expansion is multiplied by a + b, each old term produces one contribution through multiplication by a and another through multiplication by b. Interior powers in the new expansion receive contributions from two neighbouring positions. Their coefficients therefore add. The first and last positions each receive only one contribution, so their coefficients remain 1.

Example — Using the fifth row

Problem
Expand (2x + 3y)⁵ using Pascal’s triangle.

  1. 1.The coefficient row for n = 5 is 1, 5, 10, 10, 5, 1. Treat 2x as the complete first term and 3y as the complete second term.
  2. 2.Write (2x)⁵ + 5(2x)⁴(3y) + 10(2x)³(3y)² + 10(2x)²(3y)³ + 5(2x)(3y)⁴ + (3y)⁵.
  3. 3.Calculate the numerical factors: 2⁵ = 32; 5 × 2⁴ × 3 = 240; 10 × 2³ × 3² = 720; 10 × 2² × 3³ = 1080; 5 × 2 × 3⁴ = 810; 3⁵ = 243.
  4. 4.The expansion is 32x⁵ + 240x⁴y + 720x³y² + 1080x²y³ + 810xy⁴ + 243y⁵.
  5. 5.The powers of x and y sum to 5 in every term because the substituted terms 2x and 3y each have degree one.
The triangle has a long history

The chapter connects this arrangement with the Indian tradition of Meru Prastara associated with Pingala, the Chinese work of Chu-shi-kie, and later European work by Stifel, Bombelli, Oughtred, and Pascal. Pascal’s treatment was published posthumously in 1665. The same coefficient pattern appears in these different mathematical traditions.

From a triangle to combinations

Building every row is useful for small powers, but inefficient if you need only the row for power 50. Combinations give a direct expression for each entry. A combination counts how many ways a specified number of positions can be chosen from a fixed set, without caring about the order of the chosen positions.

Combination notationLaTeX
n and r are integers. n! means 1 × 2 × … × n, with 0! = 1. The notation counts selections of r positions from n positions.

The coefficients at the two ends are C(n, 0) = C(n, n) = 1. There is one way to choose no positions and one way to choose them all. The other coefficients arise because the same final product can be formed in several ways. In the product (a + b)(a + b)…(a + b), with n factors, selecting b from exactly r factors and a from the others produces aⁿ⁻ʳbʳ. There are C(n, r) possible selections of those r factors.

Pascal’s triangle written with combinationsn = 0C(0, 0)n = 1C(1, 0)C(1, 1)n = 2C(2, 0)C(2, 1)C(2, 2)n = 3C(3, 0)C(3, 1)C(3, 2)C(3, 3)n = 4C(4, 0)C(4, 1)C(4, 2)C(4, 3)C(4, 4)n = 5C(5, 0)C(5, 1)C(5, 2)C(5, 3)C(5, 4)C(5, 5)
Pascal’s triangle written with combinations— Each row is labelled by its exponent n. The entries run from r = 0 to r = n. Compare the numerical entries with the combination notation.
Example — Explaining the coefficient 6

Problem
Why is the coefficient of a²b² in (a + b)⁴ equal to 6?

  1. 1.Think of four positions, one for each factor a + b. To produce a²b², choose two positions from which to take b.
  2. 2.The six selections are {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, and {3, 4}.
  3. 3.Each selection gives the same product a²b²; adding the six contributions gives 6a²b².
  4. 4.The count is C(4, 2) = 4!/(2!2!) = 6. This explains the coefficient rather than merely locating it in the triangle.

For power 7, the coefficient row can be written directly as C(7, 0), C(7, 1), …, C(7, 7). Evaluating these entries gives 1, 7, 21, 35, 35, 21, 7, 1. No earlier row is needed. In the general theorem, these combination values are called binomial coefficients.

The theorem and its notation

We can now assemble the pattern into one statement. Each expansion position chooses r copies of the second term and n − r copies of the first. The coefficient counts the possible choices. The expression below records all positions, from r = 0 through r = n.

Binomial theoremLaTeX
n is a positive integer. a and b are the complete two terms of the binomial, not necessarily single letters.

The notation Σ, read as sigma or sum, is a shorter instruction to add a sequence of expressions. The r = 0 below it tells us where to start, and n above it tells us where to stop. Substitute each integer r in turn. Because both endpoints are used, there are n + 1 positions.

The same theorem as a sumLaTeX
r is the running index of the sum. Changing its name to k changes no mathematics. A factor raised to power zero contributes 1 in the formal expansion.

The expression C(n, r)aⁿ⁻ʳbʳ is the term at position r + 1, since counting begins at r = 0. We can use it to check any position of a full expansion. Do not confuse r with the position number: the fourth position uses r = 3. Also distinguish a binomial coefficient C(n, r) from the numerical coefficient after a and b have been replaced by expressions containing numbers.

Example — Expanding a shifted variable

Problem
Expand (x + 2)⁶.

  1. 1.The coefficients are C(6, r): 1, 6, 15, 20, 15, 6, 1.
  2. 2.Use decreasing powers of x and increasing powers of 2: x⁶ + 6x⁵(2) + 15x⁴(2²) + 20x³(2³) + 15x²(2⁴) + 6x(2⁵) + 2⁶.
  3. 3.Simplify the numerical factors to obtain x⁶ + 12x⁵ + 60x⁴ + 160x³ + 240x² + 192x + 64.
  4. 4.Check x = 0: the original expression is 2⁶ = 64, and the expansion leaves the same constant. The binomial coefficient at the third position is 15, while the coefficient of x⁴ after substitution is 60.
Count powers of the complete terms

In C(n, r)aⁿ⁻ʳbʳ, the exponents of the two placeholders sum to n. After substituting a = x² and b = 3/x, the resulting powers of x will follow a different pattern. Do not assume the exponents of the final variable must still sum to n. Raise the entire substituted expression to its assigned power.

Why the formula holds for every positive integer

A few correct rows suggest a pattern, but they do not prove that the pattern continues forever. Mathematical induction supplies that proof. We verify the first case, then show that whenever the expansion works at one power, multiplication by a + b makes it work at the next power. Together, those steps carry the result through every positive integer.

The key coefficient identity is Pascal’s addition rule written with combinations. For an interior position, a selection of r items from k + 1 items either contains the last item or does not. If it contains it, choose the other r − 1 from the first k; otherwise choose all r from the first k. The two cases are disjoint and exhaustive, so their counts add.

Pascal’s coefficient identityLaTeX
k and r are integers. The end coefficients are handled separately: C(k + 1, 0) = C(k + 1, k + 1) = 1.

Let P(n) mean the binomial formula is true at power n. At n = 1 the formula gives C(1, 0)a + C(1, 1)b = a + b, so P(1) is true. Next assume P(k) is true for some positive integer k. This is the induction assumption; we are temporarily using the formula at k to establish it at k + 1.

Induction assumptionLaTeX
We now multiply this known expansion by a + b and collect the resulting powers.

Multiplication by a gives a term C(k, r)aᵏ⁺¹⁻ʳbʳ at each interior position. Multiplication by b gives a contribution to the same power from the previous old position: C(k, r − 1)aᵏ⁺¹⁻ʳbʳ. Both contributions have identical powers, so they are like terms. Their coefficient is C(k, r) + C(k, r − 1), which equals C(k + 1, r).

Collecting the two contributionsLaTeX
The two endpoints receive only one contribution each. The sum here contains the interior positions.

Replace the bracketed coefficient by C(k + 1, r). Together with the two end coefficients 1, this is exactly the required expansion at k + 1. We have proved P(k + 1) whenever P(k) holds. Starting from P(1), this establishes P(2), then P(3), and so on. Therefore the theorem holds for every positive integer n.

Example — Seeing an induction step in a small case

Problem
Show how the cube expansion produces the fourth-power coefficients.

  1. 1.Begin with a³ + 3a²b + 3ab² + b³. Multiplying by a gives a⁴ + 3a³b + 3a²b² + ab³.
  2. 2.Multiplying by b gives a³b + 3a²b² + 3ab³ + b⁴.
  3. 3.Add like terms: a⁴ + (3 + 1)a³b + (3 + 3)a²b² + (1 + 3)ab³ + b⁴.
  4. 4.The new coefficients are 1, 4, 6, 4, 1. The general proof carries out this same collection for an arbitrary power k.

Differences and useful substitutions

A difference is a sum with a negative second term. This lets us use the same theorem while tracking signs through powers of that negative term. Substituting 1 for the first term is also useful: all its powers become 1, leaving a clear connection between the coefficients and powers of the second term.

Expansion of a differenceLaTeX
This is the theorem with a = x and b = −y. The signs alternate because (−y)ʳ = (−1)ʳyʳ.
Example — An alternating expansion

Problem
Expand (x − 2y)⁵.

  1. 1.Use a = x and b = −2y. The coefficient row is 1, 5, 10, 10, 5, 1.
  2. 2.Write x⁵ + 5x⁴(−2y) + 10x³(−2y)² + 10x²(−2y)³ + 5x(−2y)⁴ + (−2y)⁵.
  3. 3.The powers of −2y give −2y, 4y², −8y³, 16y⁴, and −32y⁵.
  4. 4.The expansion is x⁵ − 10x⁴y + 40x³y² − 80x²y³ + 80xy⁴ − 32y⁵.
  5. 5.Odd powers of the negative term contribute negative signs; even powers contribute positive signs.
Expansions around 1LaTeX
The first term is 1. Its powers all equal 1, so the coefficients stand beside ascending powers of x.

Put x = 1 in the first identity. Its left side becomes 2ⁿ and every xʳ on the right becomes 1. This proves that the sum of a coefficient row is 2ⁿ. Put x = 1 in the second identity. For positive n, its left side is 0, giving an alternating coefficient sum of zero. These are exact identities obtained by substitution, not separate rules to memorise without explanation.

Coefficient sumsLaTeX
The alternating-sum identity requires positive n. The n = 0 row has sum 1 and is useful when displaying the triangle.
Example — Checking two coefficient sums

Problem
Use the row for n = 5 to verify the ordinary and alternating sums.

  1. 1.The row is 1, 5, 10, 10, 5, 1. Its ordinary sum is 1 + 5 + 10 + 10 + 5 + 1 = 32 = 2⁵.
  2. 2.Its alternating sum is 1 − 5 + 10 − 10 + 5 − 1 = 0.
  3. 3.The terms with even r sum to 1 + 10 + 5 = 16; the terms with odd r sum to 5 + 10 + 1 = 16.
  4. 4.The equality of those two totals is another way to read an alternating sum of zero.

Check your understanding

Use the pattern and its explanation together. For each question, identify whether it concerns a coefficient, an exponent, a sign, or the logic of the proof. This keeps similar-looking expressions from being confused.

Quiz

Quick check

How many positions occur in the standard expansion of (a + b)⁷?

Quick check

Which term structure belongs in the fourth position of (a + b)⁶?

Quick check

Which row follows 1, 4, 6, 4, 1 in Pascal’s triangle?

Quick check

In an induction proof, why is verifying n = 1 alone insufficient?

Quick check

What is the sign of the numerical coefficient of x²y³ in (x − y)⁵?

Quick check

What is the sum C(6, 0) + C(6, 1) + … + C(6, 6)?

Quick check

Why does C(4, 2) appear in the coefficient of a²b²?

Practice Problems

Practice Problems
  1. Construct the rows for n = 6 and n = 7 from Pascal’s adjacent-addition rule.
  2. Expand (a + b)⁵ and explain how the exponents change between neighbouring positions.
  3. Calculate C(7, 3) using factorials and interpret it as a count of selections.
  4. Expand (2x + y)⁴, keeping the complete terms in brackets until their powers are evaluated.
  5. Explain why the coefficient of a²b³ at power 5 is C(4, 3) + C(4, 2).
  6. Expand (1 − 3x)⁴ and explain each sign.
  7. Write (x + 2)⁶ using summation notation, then identify the position with r = 2.
  8. Use substitutions in (1 + x)⁷ to find the ordinary and alternating coefficient sums.

Key Takeaways

Key Takeaways

• The standard expansion of power n has n + 1 positions with complementary exponents of the two binomial terms. • Pascal’s triangle builds each interior coefficient by adding the two adjacent entries above it. • The coefficient C(n, r) counts the choices that create aⁿ⁻ʳbʳ. • The induction proof combines contributions from two neighbouring positions to establish the next power. • A difference uses the same theorem with a negative second term, producing alternating signs. • Substitution into (1 + x)ⁿ gives ordinary, alternating, and other weighted coefficient sums.

Previous

Start of chapter

Next · Lesson 2

Expanding and Applying Binomials