Pascal's triangle and finite binomial expansion | Year 11

01 / LEARN

The idea, step by step

Understand where binomial coefficients come from

When multiplying n copies of (a+b), each term comes from choosing either a or b from every bracket. To obtain a term with r copies of b, choose the r brackets that contribute b. There are n choose r such choices. This explains both Pascal's triangle and the coefficients in a finite binomial expansion.

{(a + b)}^{n} = \sum_{r = 0}^{n} (\begin{matrix} n \\ r \end{matrix}) a^{n - r} b^{r}

Track powers, signs and the requested term

Start with descending powers of the first term and ascending powers of the second. If the second term is negative, its power controls the sign. For an expression such as (2x−3)^n, both the numerical factor and the variable are raised to the chosen power. To find one coefficient, identify the correct value of r instead of expanding everything.

A common trap

Applying the coefficient to the variable but not to its numerical multiplier.

Instead: The whole bracketed term is raised to its power: ${(2 x)}^{3} = 8 x^{3}$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

Expand with Pascal coefficients

Expand ${(x + 2)}^{4}$ .

The coefficients are $1, 4, 6, 4, 1$ .
Use $x^{4} + 4 x^{3} (2) + 6 x^{2} (2^{2}) + 4 x (2^{3}) + 2^{4}$ .
Simplify each term.

x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16

Example 2

Find one coefficient

Find the coefficient of $x^{2}$ in ${(2 x - 1)}^{5}$ .

Two factors must contribute $2 x$ and three must contribute $- 1$ .
$(\begin{matrix} 5 \\ 2 \end{matrix}) {(2)}^{2} {(- 1)}^{3} = 10 \times 4 \times (- 1) = - 40$ .

−40.

Check your understanding

What is the middle coefficient in ${(a + b)}^{4}$ ?

Show answer

The coefficient is $(\begin{matrix} 4 \\ 2 \end{matrix}) = 6$ .

03 / WORKSHEET

Your turn

6 questions · 13 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

Write the coefficients of ${(a + b)}^{5}$ .
2 marks
Expand ${(x + 1)}^{3}$ .
2 marks
Expand ${(x - 2)}^{3}$ .
2 marks
Find the coefficient of $x^{3}$ in ${(x + 2)}^{5}$ .
2 marks
Find the constant term in ${(3 x - 2)}^{4}$ .
2 marks
Show that the sum of coefficients of ${(2 x + 1)}^{4}$ is 81.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Write the coefficients of ${(a + b)}^{5}$ .
2 marks
Each interior Pascal entry is the sum of the two entries above it.
1, 5, 10, 10, 5, 1.
Expand ${(x + 1)}^{3}$ .
2 marks
Use the Pascal row 1, 3, 3, 1.
$x^{3} + 3 x^{2} + 3 x + 1$ .
Expand ${(x - 2)}^{3}$ .
2 marks
$x^{3} + 3 x^{2} (- 2) + 3 x {(- 2)}^{2} + {(- 2)}^{3}$ .
$x^{3} - 6 x^{2} + 12 x - 8$ .
Find the coefficient of $x^{3}$ in ${(x + 2)}^{5}$ .
2 marks
Choose two brackets contributing 2: $(\begin{matrix} 5 \\ 2 \end{matrix}) 2^{2} = 40$ .
40.
Find the constant term in ${(3 x - 2)}^{4}$ .
2 marks
The only constant term takes $- 2$ from all four brackets: ${(- 2)}^{4} = 16$ .
16.
Show that the sum of coefficients of ${(2 x + 1)}^{4}$ is 81.
3 marks
Setting x to 1 adds the coefficients because every power of 1 is 1.
Substitute x = 1: $3^{4} = 81$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: counting, binomial expansion, data, samples, summary measures and introductory normal distributions. These lessons provide the Counting and Statistics strand of the approved Methods preparation route.

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