Pascal's triangle and the binomial expansion – Combinatorics (Senior 4 Mathematics) | Ganzaa

Combinatorics

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Senior 4 · Mathematics

Combinatorics

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Lesson 5 of 5

Pascal's triangle and the binomial expansion

25 min

Where do the coefficients come from?

You know that $(a + b)^{2} = a^{2} + 2 a b + b^{2}$ . Multiplying once more by $(a + b)$ gives $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ . The coefficients 1, 2, 1 and 1, 3, 3, 1 are counts. When you expand

(a + b)^{3} = (a + b) (a + b) (a + b),

every term is made by picking $a$ or $b$ from each of the three brackets. The term $a^{2} b$ comes from picking $b$ from exactly one bracket, which can be done in $\binom{3}{1} = 3$ ways; so its coefficient is 3. In general, the coefficient of $a^{n-r}b^r$ in $(a + b)^{n}$ is the number of ways to choose the $r$ brackets that give $b$ , namely $\binom{n}{r}$ . Combinations are the coefficients of algebra.

Definition

Pascal's triangle: the triangular array whose row $n$ (counting from $n = 0$ ) lists the numbers $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$ . Each row begins and ends with 1, and every other entry is the sum of the two entries just above it. This is Pascal's rule, $\binom{n}{r} = \binom{n - 1}{r - 1} + \binom{n - 1}{r}$ .

Example: row 4 is 1, 4, 6, 4, 1, so $\binom{4}{2} = 6$ .

The triangle carries several patterns you already know: every row is symmetric ( $\binom{n}{r} = \binom{n}{n - r}$ ); the entries of row $n$ add up to $2^{n}$ (row 5: $1 + 5 + 10 + 10 + 5 + 1 = 32$ ); and the second diagonal is the natural numbers 1, 2, 3, 4, …

The same triangle in Zhu Shijie's Jade Mirror of the Four Unknowns, printed in China in 1303, with the numbers in rod numerals down to the row 1, 8, 28, 56, 70, 56, 28, 8, 1. In China it is called the Jia Xian or Yang Hui triangle; it was in use centuries before Blaise Pascal wrote about it in the 1650s. Public domain, via Wikimedia Commons.

Important

The binomial theorem. For any numbers $a, b$ and natural number $n$ ,

(a + b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + \dots + b^n .

The general term (the $(r + 1)$ th term) is

T_{r+1} = \binom{n}{r} a^{n-r} b^r, \qquad r = 0, 1, \dots, n .

Features: there are $n + 1$ terms; the power of $a$ falls from $n$ to 0 while the power of $b$ rises from 0 to $n$ ; in every term the powers add up to $n$ .

Worked example 1: a full expansion

Expand $(2 x - 3)^{4}$ .

Solution

Use row 4 (1, 4, 6, 4, 1) with $a = 2 x$ and $b = - 3$ . Keep the minus sign inside $b$ .

\begin{aligned} (2x - 3)^4 &= (2x)^4 + 4(2x)^3(-3) + 6(2x)^2(-3)^2 + 4(2x)(-3)^3 + (-3)^4 \\ &= 16x^4 - 96x^3 + 216x^2 - 216x + 81 . \end{aligned}

Check: put $x = 1$ : the left side is $(- 1)^{4} = 1$ , and the right side is $16 - 96 + 216 - 216 + 81 = 1$ . ✓

Worked example 2: one coefficient

Find the coefficient of $x^{5}$ in the expansion of $(1 + 2 x)^{8}$ .

Solution

$T_{r+1} = \binom{8}{r}(1)^{8-r}(2x)^r = \binom{8}{r}2^r x^r$ . For $x^{5}$ take $r = 5$ :

\binom{8}{5} \times 2^5 = 56 \times 32 = 1792 .

Worked example 3: the term independent of $x$

Find the term independent of $x$ (the constant term) in $\left(x^2 + \frac{1}{x}\right)^6$ .

Solution

T_{r+1} = \binom{6}{r}\left(x^2\right)^{6-r}\left(x^{-1}\right)^r = \binom{6}{r}x^{12 - 2r - r} = \binom{6}{r}x^{12 - 3r} .

The term is constant when $12 - 3 r = 0$ , that is $r = 4$ . It is the 5th term, $T_5 = \binom{6}{4} = 15$ .

Worked example 4: an approximation

A savings group's fund grows by 2% a month, so after 8 months it is multiplied by ${1.02}^{8}$ . Use the first four terms of a binomial expansion to estimate ${1.02}^{8}$ .

Solution

(1 + 0.02)^8 \approx 1 + 8(0.02) + 28(0.02)^2 + 56(0.02)^3 = 1 + 0.16 + 0.0112 + 0.000448 = 1.171648 .

A calculator gives $1.02^8 = 1.171659\ldots$ , so four terms are already correct to four decimal places: the fund grows by about 17.17%. The later terms are tiny because each contains a higher power of 0.02.

Warning

Three common slips. (1) Losing the sign: in $(2 x - 3)^{4}$ , $b = - 3$ , and odd powers of $b$ are negative. (2) Forgetting to raise the coefficient: $(2 x)^{3} = 8 x^{3}$ , not $2 x^{3}$ . (3) Mixing up the term number: $T_{r+1}$ contains $\binom{n}{r}$ , so the 5th term uses $r = 4$ .

Tip

For small powers, write out the row of Pascal's triangle you need by adding pairs from the row above; for large $n$ or a single term, use the general term $T_{r+1}$ . To find a particular power of $x$ , simplify the general term to one power of $x$ first, then solve for $r$ .

Mario's Math Tutoring: Tips on Expanding a Binomial Using Pascal's Triangle. Watch how the row is chosen and how the powers of each part rise and fall.

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Practice

Try each one before opening the answers. They get harder as you go.

Write row 8 of Pascal's triangle, using row 7.
Expand $(x + 2)^{5}$ .
Expand $(1 - 3 y)^{4}$ .
Find the 4th term of $(2 a + b)^{7}$ .
Find the coefficient of $x^{3}$ in $(3 - x)^{6}$ .
Find the term independent of $x$ in $\left(2x - \frac{1}{x}\right)^6$ .
Use the binomial expansion to evaluate ${0.98}^{5}$ to four decimal places.

Show answersHide answers

1, 8, 28, 56, 70, 56, 28, 8, 1.
$x^{5} + 10 x^{4} + 40 x^{3} + 80 x^{2} + 80 x + 32$ .
$1 - 12 y + 54 y^{2} - 108 y^{3} + 81 y^{4}$ .
$T_4 = \binom{7}{3}(2a)^4b^3 = 35 \times 16\,a^4b^3 = 560a^4b^3$ .
$T_{r+1} = \binom{6}{r}3^{6-r}(-x)^r$ ; $r = 3$ : $20 \times 27 \times (-1) = -540$ .
$T_{r+1} = \binom{6}{r}(2x)^{6-r}(-x^{-1})^r = \binom{6}{r}2^{6-r}(-1)^r x^{6 - 2r}$ ; $r = 3$ : $20 \times 8 \times (-1) = -160$ .
$(1 - 0.02)^5 \approx 1 - 0.1 + 0.004 - 0.00008 + 0.0000008 = 0.9039$ (to four decimal places).

Quick check

Key points

Row $n$ of Pascal's triangle is $\binom{n}{0}, \dots, \binom{n}{n}$ ; each entry is the sum of the two above.
$(a + b)^n = \sum_{r=0}^{n} \binom{n}{r}a^{n-r}b^r$ , with $n + 1$ terms.
General term $T_{r+1} = \binom{n}{r}a^{n-r}b^r$ ; set the power of $x$ to find a particular term.
Keep signs inside $b$ , and raise the whole bracket, coefficient included.

1Where do the coefficients come from?

Where do the coefficients come from?