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19.4 The Binomial Theorem

Properties of ${(a + b)}^{n}$ ${(a + b)}^{n}$ • Factorial Notation • Binomial Theorem • Pascal’s Triangle • Binomial Series

To expand ${(x + 2)}^{5}$ ${(x + 2)}^{5}$ would require a number of multiplications that would be a tedious operation. We now develop the binomial theorem, by which it is possible to expand binomials to any given power without direct multiplication. Using this type of expansion, it is also possible to expand some expressions for which direct multiplication is not possible. The binomial theorem is used to develop expressions needed in certain mathematics topics and in technical applications.

By direct multiplication, we may obtain the following expansions of the binomial $a + b :$ $a + b :$

\begin{array}{l} {(a + b)}^{0} = 1 \\ {(a + b)}^{1} = a + b \\ {(a + b)}^{2} = a^{2} + 2 a b + b^{2} \\ {(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} \\ {(a + b)}^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4} \\ {(a + b)}^{5} = a^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 a b^{4} + b^{5} \end{array}

$\begin{array}{l} {(a + b)}^{0} = 1 \\ {(a + b)}^{1} = a + b \\ {(a + b)}^{2} = a^{2} + 2 a b + b^{2} \\ {(a + b)}^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3} \\ {(a + b)}^{4} = a^{4} + 4 a^{3} b + 6 a^{2} b^{2} + 4 a b^{3} + b^{4} \\ {(a + b)}^{5} = a^{5} + 5 a^{4} b + 10 a^{3} b^{2} + 10 a^{2} b^{3} + 5 a b^{4} + b^{5} \end{array}$

Inspection shows these expansions have certain properties, and we assume that these properties are valid for the expansion of ${(a + b)}^{n},$ ${(a + b)}^{n},$ where n is any positive integer.

The following example illustrates the use of the basic properties in expanding a binomial. Carefully note the diagram in which each of these five properties is specifically noted.

EXAMPLE 1 Using basic binomial properties

Using the basic properties, develop the expansion for ${(a + b)}^{5} .$ ${(a + b)}^{5} .$

Because the exponent of the binomial is 5, we have $n = 5.$ $n = 5.$

From property 1, we know that there are six terms.

From property 2, we know that the first term is $a^{5}$ $a^{5}$ and the final term is $b^{5} .$ $b^{5} .$

From property 3, we know that the factors of a and b in terms 2, 3, 4, and 5 are $a^{4} b, a^{3} b^{2}, a^{2} b^{3},$ $a^{4} b, a^{3} b^{2}, a^{2} b^{3},$ and $a b^{4},$ $a b^{4},$ respectively.

From property 4, we obtain the coefficients of terms 2, 3, 4, and 5. In the first term, $a^{5},$ $a^{5},$ the coefficient is 1. Multiplying by 5, the power of a, and dividing by 1, the number of the term, we obtain 5, which is the coefficient of the second term. Thus, the second term is $5 a^{4} b .$ $5 a^{4} b .$ Again using property 4, we obtain the coefficient of the third term. The coefficient of the second term is 5. Multiplying by 4, and dividing by 2, we obtain 10. This means that the third term is $10 a^{3} b^{2} .$ $10 a^{3} b^{2} .$

From property 5, we know that the coefficient of the fifth term is the same as the second and that the coefficient of the fourth term is the same as the third. These properties are illustrated in the following diagram:

.4-2 Full Alternative Text

It is not necessary to use the above properties directly to expand a given binomial. If they are applied to ${(a + b)}^{n},$ ${(a + b)}^{n},$ a general formula for the expansion of a binomial may be obtained. In developing and stating the general formula, it is convenient to use the factorial notation n!, where

n! = n (n - 1) (n - 2) \dots (2) (1)

$n! = n (n - 1) (n - 2) \dots (2) (1)$ (19.8)

We see that n!, read “n factorial,” represents the product of the first n positive integers.

EXAMPLE 2 Evaluating factorials

$3! = (3) (2) (1) = 6$ $3! = (3) (2) (1) = 6$
$5! = (5) (4) (3) (2) (1) = 120$ $5! = (5) (4) (3) (2) (1) = 120$
$8! = (8) (7) (6) (5) (4) (3) (2) (1) = 40, 320$ $8! = (8) (7) (6) (5) (4) (3) (2) (1) = 40, 320$
$3! + 5! = 6 + 120 = 126$ $3! + 5! = 6 + 120 = 126$
$\frac{4!}{2!} = \frac{(4) (3) (2) (1)}{(2) (1)} = 12$ $\frac{4!}{2!} = \frac{(4) (3) (2) (1)}{(2) (1)} = 12$

In evaluating factorials, we must remember that they represent products of numbers.

CAUTION

In parts (d) and (e), we see that $3! + 5!$ $3! + 5!$ is not $(3 + 5)!$ $(3 + 5)!$ and 4!/2! is not 2!.

The evaluation of the factorials in Example 2 are shown in the calculator display in Fig. 19.10.

A calculator screen. — Fig. 19.10

Graphing calculator keystrokes: bit.ly/2NDSqb6

Figure 19.10 Full Alternative Text

THE BINOMIAL FORMULA

Based on the binomial properties, the binomial theorem states that the following binomial formula is valid for all positive integer values of n (the binomial theorem is proven through advanced methods).

{(a + b)}^{n} = a^{n} + n a^{n - 1} b + \frac{n (n - 1)}{2!} a^{n - 2} b^{2} + \frac{n (n - 1) (n - 2)}{3!} a^{n - 3} b^{3} + \dots + b^{n}

${(a + b)}^{n} = a^{n} + n a^{n - 1} b + \frac{n (n - 1)}{2!} a^{n - 2} b^{2} + \frac{n (n - 1) (n - 2)}{3!} a^{n - 3} b^{3} + \dots + b^{n}$ (19.9)

EXAMPLE 3 Using the binomial formula

Using the binomial formula, expand ${(2 x + 3)}^{6} .$ ${(2 x + 3)}^{6} .$

In using the binomial formula for ${(2 x + 3)}^{6},$ ${(2 x + 3)}^{6},$ we use 2x for a, 3 for b, and 6 for n.

Thus,

.4-2 Full Alternative Text

PASCAL’S TRIANGLE

For the first few integer powers of a binomial $a + b,$ $a + b,$ the coefficients can be obtained by setting them up in the following pattern, known as Pascal’s triangle.

We note that the first and last coefficients shown in each row are 1, and the second and next-to-last coefficients are equal to n. Other coefficients are obtained by adding the two nearest coefficients in the row above, as illustrated in Fig. 19.11 for the indicated section of Pascal’s triangle. This pattern may be continued indefinitely, although use of Pascal’s triangle is cumbersome for high values of n.

Figure 19.11 Full Alternative Text

EXAMPLE 4 Using Pascal‘s triangle

Using Pascal’s triangle, expand ${(5 s - 2 t)}^{4} .$ ${(5 s - 2 t)}^{4} .$

Here, we note that $n = 4.$ $n = 4.$ Thus, the coefficients of the five terms are 1, 4, 6, 4, and 1, respectively. Also, here we use 5s for a and $- 2 t$ $- 2 t$ for b. We are expanding this expression as ${[(5 s) + (- 2 t)]}^{4} .$ ${[(5 s) + (- 2 t)]}^{4} .$ Therefore,

An illustration displays graphing calculator with four strokes with their respective values in the next row as follows. For 4 C 0 value is 1, 4 C 1 it is 4, 4 C 2 it is 6, and 4 C 3 it is 4 and the last value 4 is highlighted.

The coefficients 1, 4, 6, 4, and 1 are called binomial coefficients. In addition to using Pascal’s triangle, these coefficients can be found on a calculator as shown in Fig. 19.12.

Figure 19.12 Full Alternative Text

In certain uses of a binomial expansion, it is not necessary to obtain all terms. Only the first few terms are required. The following example illustrates finding the first four terms of an expansion.

EXAMPLE 5 Using the binomial formula

Find the first four terms of the expansion of ${(x + 7)}^{12} .$ ${(x + 7)}^{12} .$

Here, we use x for a, 7 for b, and 12 for n. Thus, from the binomial formula, we have

\begin{aligned} {(x + 7)}^{12} & = & x^{12} + 12 x^{11} (7) + \frac{(12) (11)}{2} x^{10} (7^{2}) + \frac{(12) (11) (10)}{(2) (3)} x^{9} (7^{3}) + \dots \\ = & x^{12} + 84 x^{11} + 3234 x^{10} + 75, 460 x^{9} + \dots \end{aligned}

$\begin{aligned} {(x + 7)}^{12} & = & x^{12} + 12 x^{11} (7) + \frac{(12) (11)}{2} x^{10} (7^{2}) + \frac{(12) (11) (10)}{(2) (3)} x^{9} (7^{3}) + \dots \\ = & x^{12} + 84 x^{11} + 3234 x^{10} + 75, 460 x^{9} + \dots \end{aligned}$

If we let $a = 1$ $a = 1$ and $b = x$ $b = x$ in the binomial formula, we obtain the binomial series.

\begin{array}{l} Binomial Series \\ {(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots \end{array}

$\begin{array}{l} Binomial Series \\ {(1 + x)}^{n} = 1 + n x + \frac{n (n - 1)}{2!} x^{2} + \frac{n (n - 1) (n - 2)}{3!} x^{3} + \dots \end{array}$ (19.10)

Through advanced methods, the binomial series can be shown to be valid for any real number n if $| x | < 1.$ $| x | < 1.$ When n is either negative or a fraction, we obtain an infinite series. In such a case, we calculate as many terms as may be needed although such a series is not obtainable through direct multiplication. The binomial series may be used to develop important expressions that are used in applications and more advanced mathematics topics.

EXAMPLE 6 Binomial series—forces on a beam

In the analysis of forces on beams, the expression $1 / {(1 + m^{2})}^{3/2}$ $1 / {(1 + m^{2})}^{3/2}$ is used. Use the binomial series to find the first four terms of the expansion.

Using negative exponents, we have

1 / {(1 + m^{2})}^{3/2} = {(1 + m^{2})}^{- 3/2}

$1 / {(1 + m^{2})}^{3/2} = {(1 + m^{2})}^{- 3/2}$

Now, in using Eq. (19.10), we have $n = - 3/2$ $n = - 3/2$ and $x = m^{2} :$ $x = m^{2} :$

{(1 + m^{2})}^{- 3/2} = 1 + (- \frac{3}{2}) (m^{2}) + \frac{(- \frac{3}{2}) (- \frac{3}{2} - 1)}{2!} {(m^{2})}^{2} + \frac{(- \frac{3}{2}) (- \frac{3}{2} - 1) (- \frac{3}{2} - 2)}{3!} {(m^{2})}^{3} + \dots

${(1 + m^{2})}^{- 3/2} = 1 + (- \frac{3}{2}) (m^{2}) + \frac{(- \frac{3}{2}) (- \frac{3}{2} - 1)}{2!} {(m^{2})}^{2} + \frac{(- \frac{3}{2}) (- \frac{3}{2} - 1) (- \frac{3}{2} - 2)}{3!} {(m^{2})}^{3} + \dots$

Therefore,

\frac{1}{{(1 + m^{2})}^{3/2}} = 1 - \frac{3}{2} m^{2} + \frac{15}{8} m^{4} - \frac{35}{16} m^{6} + \dots

$\frac{1}{{(1 + m^{2})}^{3/2}} = 1 - \frac{3}{2} m^{2} + \frac{15}{8} m^{4} - \frac{35}{16} m^{6} + \dots$

EXAMPLE 7 Evaluation using binomial series

Approximate the value of ${0.97}^{7}$ ${0.97}^{7}$ by use of the binomial series.

We note that $0.97 = 1 - 0.03,$ $0.97 = 1 - 0.03,$ which means ${0.97}^{7} = {[1 + (- 0.03)]}^{7} .$ ${0.97}^{7} = {[1 + (- 0.03)]}^{7} .$ Using four terms of the binomial series, we have

\begin{aligned} {0.97}^{7} & = & {[1 + (- 0.03)]}^{7} \\ = & 1 + 7 (- 0.03) + \frac{7 (6)}{2!} {(- 0.03)}^{2} + \frac{7 (6) (5)}{3!} {(- 0.03)}^{3} \\ = & 1 - 0.21 + 0.0189 - 0.000945 = 0.807955 \end{aligned}

$\begin{aligned} {0.97}^{7} & = & {[1 + (- 0.03)]}^{7} \\ = & 1 + 7 (- 0.03) + \frac{7 (6)}{2!} {(- 0.03)}^{2} + \frac{7 (6) (5)}{3!} {(- 0.03)}^{3} \\ = & 1 - 0.21 + 0.0189 - 0.000945 = 0.807955 \end{aligned}$

From a calculator, we find that ${0.97}^{7} = 0.807983$ ${0.97}^{7} = 0.807983$ (to six decimal places), which means these values agree to four significant digits with a value 0.8080. Greater accuracy can be found using the binomial series if more terms are used.

EXERCISES 19.4

In Exercises 1 and 2, make the given changes in the indicated examples of this section and then solve the given problems.

In Example 3, change the exponent from 6 to 5.
In Example 7, change 0.97 to 0.98.

In Exercises 3–12, expand and simplify the given expressions by use of the binomial formula.

${(t + 4)}^{3}$ ${(t + 4)}^{3}$
${(x - 2)}^{3}$ ${(x - 2)}^{3}$
${(2 x - 3)}^{4}$ ${(2 x - 3)}^{4}$
${(4 x^{2} + 5)}^{4}$ ${(4 x^{2} + 5)}^{4}$
${(6 + 0.1)}^{5}$ ${(6 + 0.1)}^{5}$
${(x y^{2} - z)}^{5}$ ${(x y^{2} - z)}^{5}$
${(n + 2 π^{3})}^{5}$ ${(n + 2 π^{3})}^{5}$
${(1 - j)}^{6} (j = \sqrt{- 1})$ ${(1 - j)}^{6} (j = \sqrt{- 1})$
${(2 a - b^{2})}^{6}$ ${(2 a - b^{2})}^{6}$
${(\frac{a}{x} + x)}^{6}$ ${(\frac{a}{x} + x)}^{6}$

In Exercises 13–16, expand and simplify the given expressions by use of Pascal’s triangle.

${(5 x - 3)}^{4}$ ${(5 x - 3)}^{4}$
${(x - 4)}^{5}$ ${(x - 4)}^{5}$
${(2 a + 1)}^{6}$ ${(2 a + 1)}^{6}$
${(x - 3)}^{7}$ ${(x - 3)}^{7}$

In Exercises 17–24, find the first four terms of the indicated expansions.

${(x + 2)}^{10}$ ${(x + 2)}^{10}$
${(x - 4)}^{8}$ ${(x - 4)}^{8}$
${(2 a - 1)}^{7}$ ${(2 a - 1)}^{7}$
${(3 b + 2)}^{9}$ ${(3 b + 2)}^{9}$
${(x^{1/2} - 4 y)}^{12}$ ${(x^{1/2} - 4 y)}^{12}$
${(2 a - x^{- 1})}^{11}$ ${(2 a - x^{- 1})}^{11}$
${(b^{2} + \frac{1}{2 b})}^{20}$ ${(b^{2} + \frac{1}{2 b})}^{20}$
${(2 x^{2} + \frac{y}{3})}^{15}$ ${(2 x^{2} + \frac{y}{3})}^{15}$

In Exercises 25–28, approximate the value of the given expression to three decimal places by using three terms of the appropriate binomial series. Check using a calculator.

${1.05}^{6}$ ${1.05}^{6}$
$\sqrt{0.927}$ $\sqrt{0.927}$
$\sqrt[3]{1.045}$ $\sqrt[3]{1.045}$
${0.98}^{- 7}$ ${0.98}^{- 7}$

In Exercises 29–36, find the first four terms of the indicated expansions by use of the binomial series.

${(1 + x)}^{8}$ ${(1 + x)}^{8}$
${(1 + x)}^{- 1/3}$ ${(1 + x)}^{- 1/3}$
${(1 - 3 x)}^{- 2}$ ${(1 - 3 x)}^{- 2}$
${(1 - 2 \sqrt{x})}^{9}$ ${(1 - 2 \sqrt{x})}^{9}$
$\sqrt{1 + x}$ $\sqrt{1 + x}$
$\frac{1}{\sqrt{1 - x}}$ $\frac{1}{\sqrt{1 - x}}$
$\frac{1}{\sqrt{9 - 9 x}}$ $\frac{1}{\sqrt{9 - 9 x}}$
$\sqrt{4 + x^{2}}$ $\sqrt{4 + x^{2}}$

In Exercises 37–40, solve the given problems involving factorials.

Using a calculator, evaluate (a) $17! + 4!,$ $17! + 4!,$ (b) 21!, (c) $17! \times 4!,$ $17! \times 4!,$ and (d) 68!.
Using a calculator, evaluate (a) $8! - 7!,$ $8! - 7!,$ (b) 8!/7!, (c) $8! \times 7!$ $8! \times 7!$ , and (d) 56!.
Show that $n! = n \times (n - 1)!$ $n! = n \times (n - 1)!$ for $n \geq 2.$ $n \geq 2.$ To use this equation for $n = 1,$ $n = 1,$ explain why it is necessary to define $0! = 1$ $0! = 1$ (this is a standard definition of 0!).
Show that $\frac{(n + 1)!}{(n - 2)!} = n^{3} - n$ $\frac{(n + 1)!}{(n - 2)!} = n^{3} - n$ for $n \geq 2.$ $n \geq 2.$ See Exercise 39.

In Exercises 41–44, find the indicated terms by use of the following information. The $r + 1$ $r + 1$ term of the expansion of ${(a + b)}^{n}$ ${(a + b)}^{n}$ is given by

\overset{r factors}{\overset{︷}{\frac{n (n - 1) (n - 2) \dots (n - r + 1)}{r!}}} a^{n - r} b^{r}

$\overset{r factors}{\overset{︷}{\frac{n (n - 1) (n - 2) \dots (n - r + 1)}{r!}}} a^{n - r} b^{r}$

The term involving $b^{5}$ $b^{5}$ in ${(a + b)}^{8}$ ${(a + b)}^{8}$
The term involving $y^{6}$ $y^{6}$ in ${(x + y)}^{10}$ ${(x + y)}^{10}$
The fifth term of ${(2 x - 3 b)}^{12}$ ${(2 x - 3 b)}^{12}$
The sixth term of ${(\sqrt{a} - \sqrt{b})}^{14}$ ${(\sqrt{a} - \sqrt{b})}^{14}$

In Exercises 45–58, solve the given problems.

Without expanding, evaluate

${(2 x - 1)}^{3} + 3 {(2 x - 1)}^{2} (3 - 2 x) + 3 (2 x - 1) {(3 - 2 x)}^{2} + {(3 - 2 x)}^{3}$ ${(2 x - 1)}^{3} + 3 {(2 x - 1)}^{2} (3 - 2 x) + 3 (2 x - 1) {(3 - 2 x)}^{2} + {(3 - 2 x)}^{3}$
Add up the coefficients of the first few powers of ${(a + b)}^{n}$ ${(a + b)}^{n}$ by using Pascal’s triangle. Then by noting the pattern, determine the sum of the coefficients of ${(a + b)}^{79} .$ ${(a + b)}^{79} .$
Explain why n! ends in a zero if $n > 4.$ $n > 4.$
Expand ${[(a + b) + c]}^{3}$ ${[(a + b) + c]}^{3}$ using the binomial theorem. Group terms as indicated.
A mechanical system has 4 redundant engines, each with a 95% probability of functioning properly. When the binomial theorem is used to expand ${(0.95 + 0.05)}^{4},$ ${(0.95 + 0.05)}^{4},$ the resulting terms give the probabilities of exactly 4, 3, 2, 1, and 0 engines functioning properly. Find the probability of exactly 3 engines functioning properly (the term containing ${0.95}^{3}$ ${0.95}^{3}$ ). Round to the third decimal place.
In Exercise 49, find the probability that at least one engine functions properly (the sum of the terms containing ${0.95}^{4}, {0.95}^{3}, {0.95}^{2},$ ${0.95}^{4}, {0.95}^{3}, {0.95}^{2},$ and ${0.95}^{1}$ ${0.95}^{1}$ ). Round to the third decimal place.
Approximate $\sqrt{6}$ $\sqrt{6}$ to hundredths by noting that $\sqrt{6} = \sqrt{4 (1.5)} = 2 \sqrt{1 + 0.5}$ $\sqrt{6} = \sqrt{4 (1.5)} = 2 \sqrt{1 + 0.5}$ and using four terms of the appropriate binomial series.
Approximate $\sqrt[3]{10}$ $\sqrt[3]{10}$ by using the method of Exercise 51.
A company purchases a piece of equipment for A dollars, and the equipment depreciates at a rate of r each year. Its value V after n years is $V = A {(1 - r)}^{n} .$ $V = A {(1 - r)}^{n} .$ Expand this expression for $n = 5.$ $n = 5.$
In finding the rate of change of emission of energy from the surface of a body at temperature T, the expression ${(T + h)}^{4}$ ${(T + h)}^{4}$ is used. Expand this expression.
In the theory associated with the magnetic field due to an electric current, the expression $1 - \frac{x}{\sqrt{a^{2} + x^{2}}}$ $1 - \frac{x}{\sqrt{a^{2} + x^{2}}}$ is found. By expanding ${(a^{2} + x^{2})}^{- 1/2},$ ${(a^{2} + x^{2})}^{- 1/2},$ find the first three nonzero terms that could be used to approximate the given expression.
In the theory related to the dispersion of light, the expression $1 + \frac{A}{1 - λ_{0}^{2} / λ^{2}}$ $1 + \frac{A}{1 - λ_{0}^{2} / λ^{2}}$ arises. (a) Let $x = λ_{0}^{2} / λ^{2}$ $x = λ_{0}^{2} / λ^{2}$ and find the first four terms of the expansion of ${(1 - x)}^{- 1} .$ ${(1 - x)}^{- 1} .$ (b) Find the same expansion by using long division. (c) Write the original expression in expanded form, using the results of (a) and (b).
A cubic Bezier curve is defined by four points ( $P_{0}$ $P_{0}$ through $P_{3}$ $P_{3}$ ) from the parametric equations given by ${(1 - t)}^{3} P_{0} + 3 {(1 - t)}^{2} t P_{1} + 3 (1 - t) t^{2} P_{2} + t^{3} P_{3},$ ${(1 - t)}^{3} P_{0} + 3 {(1 - t)}^{2} t P_{1} + 3 (1 - t) t^{2} P_{2} + t^{3} P_{3},$ where $0 \leq t \leq 1.$ $0 \leq t \leq 1.$ Note that the points $P_{0}$ $P_{0}$ through $P_{3}$ $P_{3}$ are multiplied by terms of the binomial expansion ${[(1 - t) + t]}^{3} .$ ${[(1 - t) + t]}^{3} .$ Using a similar process, find the equation of the fourth-degree Bezier curve defined by the five points $P_{0}$ $P_{0}$ through $P_{4}$ $P_{4}$ (see Example 6 in Section 15.3).
Find the first four terms of the expansion of ${(1 + x)}^{- 1}$ ${(1 + x)}^{- 1}$ and then divide $1 + x$ $1 + x$ into 1. Compare the results.

Answers to Practice Exercises

72
$625 s^{4} + 500 s^{3} t + 150 s^{2} t^{2} + 20 s t^{3} + t^{4}$ $625 s^{4} + 500 s^{3} t + 150 s^{2} t^{2} + 20 s t^{3} + t^{4}$
$x^{9} - 36 x^{8} + 576 x^{7} - \dots$ $x^{9} - 36 x^{8} + 576 x^{7} - \dots$
1.230 (rounded off)

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Table of Contents for 19.4 The Binomial Theorem

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Table of Contents for
19.4 The Binomial Theorem