Grade by Grade
C. Discrete Functions · Sequences and Series

Pascal's Triangle and the Binomial Expansion

C1.6 C1.5About 75 minutes
Learning goals
  • I can explain the connection between Pascal's triangle and the expansion of (a + b)ⁿ.
  • I can expand a binomial raised to a whole-number exponent, including ones with coefficients and negative terms.
  • I can find a single term of a binomial expansion without writing out the whole expansion.

Words to know

binomial
A polynomial with two terms, such as a + b, x − 3 or 2x + 5y.
binomial expansion
The result of multiplying out a power of a binomial, such as (a + b)³ = a³ + 3a²b + 3ab² + b³.
coefficient
The number multiplying the variables in a term. In 6a²b², the coefficient is 6.
whole-number exponent
An exponent 0, 1, 2, 3, ….

Multiplying out, the long way and the short way

In Grade 10 you learned the special product (a + b)² = a² + 2ab + b². What about (x + 2)⁵? You could multiply (x + 2)(x + 2)(x + 2)(x + 2)(x + 2) step by step, collecting like terms each time. That is a lot of algebra and a lot of chances to slip.

Today you will find a shortcut, and it comes from the triangle in the last lesson.

a²ababb²abab(a + b)² = a² + 2ab + b²coefficients 1, 2, 1 = row 2of Pascal's triangle
A square of side a + b is made of an a² square, a b² square and two ab rectangles. That is why (a + b)² has a middle term 2ab.

Investigating the pattern

Here are the first few powers of (a + b), multiplied out:

  • (a + b)⁰ = 1
  • (a + b)¹ = a + b
  • (a + b)² = a² + 2ab + b²
  • (a + b)³ = a³ + 3a²b + 3ab² + b³
  • (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

Look only at the coefficients: 1 / 1 1 / 1 2 1 / 1 3 3 1 / 1 4 6 4 1. These are the rows of Pascal's triangle!

(a + b)³ = (a + b)(a² + 2ab + b²)a × (a² + 2ab + b²) =a³+ 2a²b+ ab²b × (a² + 2ab + b²) =+ a²b+ 2ab²+ b³add like terms:a³+ 3a²b+ 3ab²+ b³coefficients:11 + 2 = 32 + 1 = 31Shifting one place and adding is exactly how each row of Pascal's triangle is built.
Multiplying (a² + 2ab + b²) by (a + b): the a-row and the b-row line up one place apart, and adding them gives the coefficients 1, 3, 3, 1, just as Pascal's triangle adds the two numbers above.

Why it works

To get (a + b)ⁿ from (a + b)ⁿ⁻¹, you multiply by (a + b): once by a and once by b. The two results line up one place apart, and like terms are added. Each new coefficient is the sum of two coefficients from the row before. That is exactly the recursive rule of Pascal's triangle, tₙ,ᵣ = tₙ₋₁,ᵣ₋₁ + tₙ₋₁,ᵣ.

The rules for expanding (a + b)ⁿ

  1. There are n + 1 terms.
  2. The coefficients are the entries of row n of Pascal's triangle.
  3. The powers of a go down from aⁿ to a⁰ = 1.
  4. The powers of b go up from b⁰ = 1 to bⁿ.
  5. In every term the exponents add up to n.

So (a + b)⁵ = a⁵ + 5a⁴b + 10a³b² + 10a²b³ + 5ab⁴ + b⁵.

Worked example

Example 1: A binomial with a number

Expand (x + 2)⁴.

  1. Row 4: 1, 4, 6, 4, 1. Here a = x and b = 2.
  2. (x + 2)⁴ = 1(x)⁴ + 4(x)³(2) + 6(x)²(2)² + 4(x)(2)³ + 1(2)⁴
  3. = x⁴ + 4x³(2) + 6x²(4) + 4x(8) + 16
  4. = x⁴ + 8x³ + 24x² + 32x + 16
  5. Check with x = 1: left side (1 + 2)⁴ = 81; right side 1 + 8 + 24 + 32 + 16 = 81 ✓.

x⁴ + 8x³ + 24x² + 32x + 16

Worked example

Example 2: Coefficients and a negative term

Expand (2x − 3)³.

  1. Row 3: 1, 3, 3, 1. Here a = 2x and b = −3 (keep the negative sign with b, and use brackets).
  2. (2x − 3)³ = (2x)³ + 3(2x)²(−3) + 3(2x)(−3)² + (−3)³
  3. = 8x³ + 3(4x²)(−3) + 3(2x)(9) + (−27)
  4. = 8x³ − 36x² + 54x − 27
  5. Check with x = 1: (2 − 3)³ = −1 and 8 − 36 + 54 − 27 = −1 ✓.

8x³ − 36x² + 54x − 27

Worked example

Example 3: Alternating signs

Expand (x − y)⁵.

  1. Row 5: 1, 5, 10, 10, 5, 1, with a = x and b = −y.
  2. Odd powers of (−y) are negative and even powers are positive, so the signs alternate +, −, +, −, +, −.
  3. (x − y)⁵ = x⁵ − 5x⁴y + 10x³y² − 10x²y³ + 5xy⁴ − y⁵

x⁵ − 5x⁴y + 10x³y² − 10x²y³ + 5xy⁴ − y⁵

Worked example

Example 4: Finding one term

Find the 4th term in the expansion of (x + 3)⁶.

  1. Row 6: 1, 6, 15, 20, 15, 6, 1.
  2. The 1st term has b⁰, the 2nd has b¹, so the 4th term has b³ and coefficient 20 (the 4th entry). The power of a is 6 − 3 = 3.
  3. 4th term = 20(x)³(3)³ = 20(27)x³ = 540x³

540x³

Worked example

Example 5: Estimating a power

Use the binomial expansion to calculate 1.01⁵.

  1. Write 1.01⁵ = (1 + 0.01)⁵. Row 5: 1, 5, 10, 10, 5, 1.
  2. = 1 + 5(0.01) + 10(0.01)² + 10(0.01)³ + 5(0.01)⁴ + (0.01)⁵
  3. = 1 + 0.05 + 0.001 + 0.00001 + 0.00000005 + 0.0000000001
  4. = 1.0510100501. The first three terms already give 1.051, accurate to three decimal places.

1.0510100501 (about 1.051)

Avoid the classic mistakes

  • (x + 2)² is not x² + 4. You need all n + 1 terms.
  • Put brackets around each part: (2x)³ = 8x³, not 2x³.
  • Keep the negative sign inside the brackets: (−3)² = 9 but (−3)³ = −27.
  • Check your answer by substituting a simple value such as x = 1 into both sides.
Just for fun

A binomial walks into a gym and says, "I'd like to expand."

The trainer says, "Great. Which row are we working on today?"

Hands-on activity

Build (a + b)³ as a cube

You need: 27 sugar cubes, dice or building blocks of the same size (or draw it on paper).

  1. Build a big 3 × 3 × 3 cube. Think of each edge as a + b with a = 2 blocks and b = 1 block.
  2. Take it apart into pieces: one 2 × 2 × 2 cube (a³), three 2 × 2 × 1 slabs (a²b), three 2 × 1 × 1 rods (ab²) and one 1 × 1 × 1 cube (b³).
  3. Count the blocks in each kind of piece. Do they add to 27?
  4. Explain how this shows (a + b)³ = a³ + 3a²b + 3ab² + b³.

What to look for: a³ = 8, 3a²b = 3 × 4 = 12, 3ab² = 3 × 2 = 6, b³ = 1, and 8 + 12 + 6 + 1 = 27 = 3³ ✓. The 1, 3, 3, 1 counts of the pieces are row 3 of Pascal's triangle. (If you use sugar cubes, ask before eating any!)

Your turn

1

Expand (a + b)⁶.

2

Expand (x + 1)⁵.

3

Expand (x − 2)⁴.

4

Expand (3x + 1)³.

5

How many terms are in the expansion of (x + y)¹²?

6

Find the 3rd term of (x + 2)⁷.

7

Expand (2a − b)⁴.

More practice: warm-up

1

Which row of Pascal's triangle gives the coefficients of (a + b)³?

2

Expand (x + y)³.

3

In the expansion of (a + b)⁸, what do the exponents of each term add up to?

4

Expand (x + 3)².

5

What is the last term of (x − 1)⁷?

More practice: on your own

1

Expand (2x + 3)⁴.

2

Expand (1 − x)⁶.

3

Find the middle term of (x − 2y)⁴.

4

Expand (x² + 1)³.

5

Use the binomial expansion to find 1.02⁴ to 4 decimal places.

Challenge

1

Expand (x² − 2/x)³ and simplify each term.

2

Use the binomial expansion of (1 + 1)ⁿ and (1 − 1)ⁿ to explain two patterns from the last lesson.

3

In the expansion of (x + k)⁵, the coefficient of x³ is 90. Find all possible values of k.

Check your understanding

Quiz

1.

The coefficients of (a + b)⁴ are:

2.

How many terms are in the expansion of (2x − y)⁹?

3.

What is the expansion of (x − 1)³?

4.

What is the 2nd term of (x + 5)⁴?

5.

What is the last term of (2x + 3)⁵?

6.

Which statement explains why the coefficients of (a + b)ⁿ are the rows of Pascal's triangle?

For parents and teachers

Big idea: Expectation C1.6 asks students to discover, through investigation, the link between Pascal's triangle and the expansion of binomials, and to use it to expand binomials raised to whole-number exponents. In Grade 12 Data Management (MDM4U), the entries of Pascal's triangle return as combinations.

How to help: Ask your teen to say the five rules out loud (n + 1 terms, coefficients from row n, powers of a going down, powers of b going up, exponents adding to n). For every expansion, check by substituting x = 1 into both sides: the totals must match.

Common difficulties:

  • Forgetting brackets, especially (2x)³ = 8x³ and (−3)² = 9.
  • Losing negative signs: with (x − y)ⁿ, the signs alternate.
  • Picking the wrong term: the kth term has b to the power k − 1.
  • Writing (x + 2)² = x² + 4. Remind them of the area model.