IB Analysis and Approaches HL · Algebra review and mathematical induction · Section 6 of 6

Binomial theorem

3 worked examples · By Álvaro López

Key idea

The binomial theorem gives the expansion of $(a + b)^{n}$ without multiplying out $n$ brackets. In this section you learn to calculate the coefficients, to write the general term and to locate a specific term (the one in $x^{k}$ or the constant term), which is what the IB exam asks for. In HL, moreover, it extends to negative or fractional exponents.

Factorial

For example, $5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120$ . Factorials grow very quickly, so whole numbers are almost never calculated: they are simplified. Since $n! = n \cdot (n - 1)!$ , a quotient is reduced by cancelling the common factor:

$\frac{8 !}{6 !} = \frac{8 \cdot 7 \cdot 6 !}{6 !} = 8 \cdot 7 = 56$

The convention $0! = 1$ is not arbitrary: it makes the formulas below work in the extreme cases.

Binomial coefficients

Example: from a group of $8$ students, $3$ are chosen for a committee. The number of possible committees is

$(3 8) = \frac{8 !}{3 ! 5 !} = \frac{8 \cdot 7 \cdot 6}{3 \cdot 2 \cdot 1} = 56$

Notice the shortcut: the numerator is the product of $r$ factors counting down from $n$ , and the denominator is $r!$ . That is,

$(r n) = \frac{n ( n - 1 ) ( n - 2 ) \dots ( n - r + 1 )}{r !}$

This form is the one that is generalised at the end of the section. In Paper 1 it is calculated by hand with that shortcut; in Paper 2 you can obtain any binomial coefficient directly with the graphic display calculator (GDC).

The intuition behind each one, with a group of $n$ people:

There is only one way to choose nobody ( $r = 0$ ) and only one way to choose everybody ( $r = n$ ).
Choosing the people who get in to the committee is the same as choosing the people who stay out: that is why $(3 8) = (5 8) = 56$ . It saves work: $(10 12) = (2 12) = 66$ .
To choose $r + 1$ people from a group of $n + 1$ , fix one particular person: either you do not choose them (you still need $r + 1$ from the other $n$ ) or you do choose them (you still need $r$ from the other $n$ ). Hence the sum. Numerical check: $(2 6) + (3 6) = 15 + 20 = 35 = (3 7)$ .

Pascal's triangle

Pascal's formula lets you build binomial coefficients without calculating factorials: each number is the sum of the two above it. Row $n$ (counting from $n = 0$ ) contains $(0 n), (1 n), \dots, (n n)$ :

$111615141513101262013101415151611$

Each row is symmetric (property 2) and starts and ends with $1$ (property 1). Moreover, the sum of row $n$ is $2^{n}$ : row $5$ gives $1 + 5 + 10 + 10 + 5 + 1 = 32 = 2^{5}$ . For small $n$ (up to about $6$ or $8$ ), the triangle is the fastest way to obtain the coefficients.

The binomial theorem

Expand $(a + b)^{3}$ by multiplying it out: $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ . The coefficients $1, 3, 3, 1$ are row $3$ of the triangle, that is, $(0 3), (1 3), (2 3), (3 3)$ . This is no coincidence.

Why it works: when multiplying out $n$ brackets $(a + b) (a + b) \dots (a + b)$ , from each bracket you choose either $a$ or $b$ . To obtain $a^{n - r} b^{r}$ you must choose $b$ in exactly $r$ of the $n$ brackets, and there are $(r n)$ ways to decide which ones. That number is the coefficient.

What to look for in the expansion:

It has $n + 1$ terms.
The powers of $a$ go down from $n$ to $0$ and the powers of $b$ go up from $0$ to $n$ ; in each term the two exponents add up to $n$ .
If $b$ is negative, the signs alternate: $(a - b)^{n}$ is obtained by putting $b \to (- b)$ .
As the IB treats this formula as known in the course (it appears in the formula booklet), what is assessed is knowing how to use it: substituting $a$ and $b$ correctly and simplifying.

General term and finding a term

In the expansion of $(a + b)^{n}$ , the term containing $b^{r}$ is in position $r + 1$ (because $r$ starts at $0$ ). It is called the general term:

Almost all IB questions on the binomial are solved with the general term, not by expanding everything. What you have to do is write the general term, simplify its powers of $x$ and work out which value of $r$ gives the required exponent.

A short case: the term in $x^{2}$ of $(1 + 3 x)^{5}$ . Here $a = 1$ , $b = 3 x$ , and $T_{r + 1} = (r 5) (3 x)^{r}$ carries $x^{r}$ , so $r = 2$ and the term is $(2 5) \cdot 3^{2} x^{2} = 10 \cdot 9 x^{2} = 90 x^{2}$ .

If the expansion is multiplied by another expression, such as $(x + 2) (1 + 3 x)^{5}$ , you look separately for the terms of $(1 + 3 x)^{5}$ which, when multiplied by $x$ or by $2$ , give the required power, and add them. For the $x^{2}$ term: $x \cdot (15 x) + 2 \cdot (90 x^{2}) = 15 x^{2} + 180 x^{2} = 195 x^{2}$ , using that the $x$ term of $(1 + 3 x)^{5}$ is $(1 5) \cdot 3 x = 15 x$ .

Generalisation: negative and fractional exponents (HL only)

Until now $n$ was a non-negative integer and the expansion was finite. In HL the theorem extends to any rational exponent $n$ (negative or fractional), but then the expansion is an infinite series and is only valid for certain values of $x$ .

Notice that the coefficient of $x^{r}$ is the shortcut $\frac{n ( n - 1 ) \dots ( n - r + 1 )}{r !}$ that we saw for $(r n)$ , now with any $n$ . If $n$ is a positive integer, one of the factors $n - j$ becomes zero at some point and the series terminates: the usual theorem is recovered. If $n$ is negative or fractional, it never becomes zero and the terms continue without end.

Two well-known cases:

$(1 + x)^{- 1} = 1 - x + x^{2} - x^{3} + \dots (1 + x)^{- 2} = 1 - 2 x + 3 x^{2} - 4 x^{3} + \dots$

The first is the geometric series with ratio $- x$ , which only converges if $∣ x ∣ < 1$ : that is why this condition cannot be omitted. With $n = \frac{1}{2}$ you get $1 + x = 1 + \frac{x}{2} - \frac{x ^{2}}{8} + \frac{x ^{3}}{16} - \dots$ .

Remember that the expansion has the form $(1 + x)^{n}$ . If the binomial does not start with $1$ , you must factor out the first term:

$(a + b x)^{n} = a^{n} (1 + \frac{b}{a} x)^{n}$

and apply the series with $\frac{b}{a} x$ in place of $x$ . The validity condition then becomes $\frac{b}{a} x < 1$ , that is, $∣ x ∣ < \frac{a}{b}$ .

Worked examples

Example 1 Basic

Expanding a binomial with a negative coefficient and sign

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

Solution

Identify. $a = 2 x$ , $b = - 3$ , $n = 4$ . The coefficients are row $4$ of Pascal's triangle: $1, 4, 6, 4, 1$ .
Write each term (with $b = - 3$ in brackets, so as not to lose the signs):

$(2 x - 3)^{4} = (2 x)^{4} + 4 (2 x)^{3} (- 3) + 6 (2 x)^{2} (- 3)^{2} + 4 (2 x) (- 3)^{3} + (- 3)^{4}$

Simplify each term.

$(2 x)^{4} = 16 x^{4}$ .
$4 (2 x)^{3} (- 3) = 4 \cdot 8 x^{3} \cdot (- 3) = - 96 x^{3}$ .
$6 (2 x)^{2} (- 3)^{2} = 6 \cdot 4 x^{2} \cdot 9 = 216 x^{2}$ .
$4 (2 x) (- 3)^{3} = 4 \cdot 2 x \cdot (- 27) = - 216 x$ .
$(- 3)^{4} = 81$ .

$(2 x - 3)^{4} = 16 x^{4} - 96 x^{3} + 216 x^{2} - 216 x + 81$

Check with $x = 1$ : the original gives $(2 - 3)^{4} = 1$ , and the expansion gives $16 - 96 + 216 - 216 + 81 = 1$ .

The signs alternate because $b$ is negative: $- 3$ is raised to both even and odd powers.

Example 2 Medium

Term in x^6 and constant term

Consider the expansion of $(x^{2} - \frac{3}{x})^{9}$ .

a) Find the coefficient of the term in $x^{6}$ .

b) Find the constant term.

c) Is there a term in $x^{5}$ ? If so, state its coefficient.

Solution

General term. With $a = x^{2}$ , $b = - \frac{3}{x}$ and $n = 9$ :

$T_{r + 1} = (r 9) (x^{2})^{9 - r} (- \frac{3}{x})^{r} = (r 9) (- 3)^{r} x^{18 - 2 r} x^{- r} = (r 9) (- 3)^{r} x^{18 - 3 r}$

Exponent of $x$ : $18 - 3 r$ , with $r = 0, 1, \dots, 9$ .

a) Term in $x^{6}$ . $18 - 3 r = 6 ⟺ r = 4$ , which is an integer between $0$ and $9$ . The coefficient is

$(4 9) (- 3)^{4} = 126 \cdot 81 = 10206$

b) Constant term. $18 - 3 r = 0 ⟺ r = 6$ . The term is

$(6 9) (- 3)^{6} = 84 \cdot 729 = 61236$

c) Term in $x^{5}$ . $18 - 3 r = 5 ⟺ r = \frac{13}{3}$ , which is not an integer. There is no such term: all the exponents are multiples of $3$ ( $18, 15, 12, \dots, - 9$ ). Its coefficient is $0$ .

Note that the exponent of $x$ goes down by $3$ each time, not by $1$ : that is why some exponents never appear.

Example 3 Hard

Cube root: infinite series, validity and approximation (HL)

The function $f (x) = 3 8 + x$ is considered.

a) Write the expansion of $f (x)$ in powers of $x$ up to the term in $x^{3}$ .

b) Find the largest number $M$ such that the expansion is valid for all $x$ with $∣ x ∣ < M$ .

c) Use the terms up to $x^{2}$ of the expansion with $x = 1$ to approximate $39$ . Give the result as a fraction.

Solution

a) 1. Factor out. The binomial does not start with $1$ , so we factor out the $8$ :

$3 8 + x = (8 + x)^{1/3} = 8^{1/3} (1 + \frac{x}{8})^{1/3} = 2 (1 + \frac{x}{8})^{1/3}$

Apply the generalised theorem with $n = \frac{1}{3}$ and $\frac{x}{8}$ in place of $x$ . The coefficients are:

$x^{1}$ : $n = \frac{1}{3}$ .
$x^{2}$ : $\frac{\frac{1}{3} \cdot ( - \frac{2}{3} )}{2 !} = - \frac{1}{9}$ .
$x^{3}$ : $\frac{\frac{1}{3} \cdot ( - \frac{2}{3} ) \cdot ( - \frac{5}{3} )}{3 !} = \frac{\frac{10}{27}}{6} = \frac{5}{81}$ .

$(1 + \frac{x}{8})^{1/3} = 1 + \frac{1}{3} \cdot \frac{x}{8} - \frac{1}{9} (\frac{x}{8})^{2} + \frac{5}{81} (\frac{x}{8})^{3} + \dots = 1 + \frac{x}{24} - \frac{x ^{2}}{576} + \frac{5 x ^{3}}{41472} + \dots$

Multiply by $2$ :

$f (x) = 2 + \frac{x}{12} - \frac{x ^{2}}{288} + \frac{5 x ^{3}}{20736} + \dots$

b) The theorem requires $\frac{x}{8} < 1$ , that is, $∣ x ∣ < 8$ . The largest value is $M = 8$ .

c) With $x = 1$ (which satisfies $∣ x ∣ < 8$ ) and up to the term in $x^{2}$ :

$39 \approx 2 + \frac{1}{12} - \frac{1}{288} = \frac{576 + 24 - 1}{288} = \frac{599}{288} \approx 2.0799$

The actual value is $2.0801$ (to four decimal places): the error is about two ten-thousandths. Adding the term in $x^{3}$ would reduce it further.

In the exam: in a), marks are awarded separately for factoring out, for using $\frac{x}{8}$ (not $x$ ) in the formula and for simplifying the coefficients correctly; the most costly error is forgetting the factor $2$ at the start. Write the condition $∣ x ∣ < 8$ even if it is not explicitly asked for.

Common mistakes

Summary

$n! = n (n - 1) \dots 2 \cdot 1$ and $0! = 1$ ; quotients of factorials are simplified by cancelling.
$(r n) = \frac{n !}{r ! ( n - r )!} = \frac{n ( n - 1 ) \dots ( n - r + 1 )}{r !}$ counts the ways of choosing $r$ elements from $n$ .
Properties: $(0 n) = (n n) = 1$ , $(r n) = (n - r n)$ and $(r n) + (r + 1 n) = (r + 1 n + 1)$ . Pascal's triangle collects them: each number is the sum of the two above it.
Binomial theorem: $(a + b)^{n} = \sum_{r = 0}^{n} (r n) a^{n - r} b^{r}$ , with $n + 1$ terms; the general term is $T_{r + 1} = (r n) a^{n - r} b^{r}$ .
To find the term in $x^{k}$ (or the constant term, $k = 0$ ): exponent of $x$ as a function of $r$ , set equal to $k$ , check that $r$ is an integer from $0$ to $n$ , and calculate the coefficient.
HL only: if $n \in Q$ and $∣ x ∣ < 1$ , $(1 + x)^{n} = 1 + n x + \frac{n ( n - 1 )}{2 !} x^{2} + \dots$ is an infinite series. For $(a + b x)^{n}$ factor out $a^{n}$ ; the validity condition is $\frac{b}{a} x < 1$ .

Keep studying

Maths classes in Aldaia and online

Want to work on it together?

Free first consultation: we look at your level and your exam, and build a plan. Álvaro replies on WhatsApp.

Message us on WhatsApp