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Section 8.5 The Binomial Theorem

Before Starting this Section, Review

1 Special products (Section P.3 , page 34)
2 Factorials (Section 8.1 , page 721)
3 Summation notation (Section 8.1 , page 722)

Objectives

1 Use Pascal’s Triangle to compute binomial coefficients.
2 Use Pascal’s Triangle to expand a binomial power.
3 Use the Binomial Theorem to expand a binomial power.
4 Find the coefficient of a term in a binomial expansion.

Blaise Pascal

Blaise Pascal was a French mathematician. An evident genius, he was 14 when he began to accompany his father to gatherings of mathematicians that were arranged by Father Marin Mersenne. At the age of 16, Pascal presented his own results at one of Mersenne’s meetings.

Pascal’s desire to help his father with his work collecting taxes led him to invent the first digital calculator, called the Pascaline. Pascal was also interested in atmospheric pressure, and in 1648, he observed that the pressure of the atmosphere decreased with height and that a vacuum existed above the atmosphere.

Pascal’s intense interest in mathematics led him to produce important results related to conic sections and to engage in correspondence with Fermat in which he formulated the foundations for the theory of probability. Pascal died a painful death from cancer at the age of 39.

In this section, we study Pascal’s Triangle. This triangle of numbers contains the important binomial coefficients. See page 760. Pascal’s work was influential in Newton’s discovery of the general Binomial Theorem for fractional and negative powers.

Binomial Expansions

Recall that a polynomial that has exactly two terms is called a binomial. In the current section, we study a method for expanding (x+y)n ${(x + y)}^{n}$ for any positive integer n.

First, consider these binomial expansions:

(x + y) 1 (x + y) 2 (x + y) 3 (x + y) 4 (x + y) 5 = = = = = x + y x 2 + 2 x y + y 2 x 3 + 3 x 2 y + 3 x y 2 + y 3 x 4 + 4 x 3 y + 6 x 2 y 2 + 4 x y 3 + y 4 x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 x y 4 + y 5

$\begin{array}{rcl} {(x + y)}^{1} & = & x + y \\ {(x + y)}^{2} & = & x^{2} + 2 x y + y^{2} \\ {(x + y)}^{3} & = & x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} \\ {(x + y)}^{4} & = & x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + y^{4} \\ {(x + y)}^{5} & = & x^{5} + 5 x^{4} y + 10 x^{3} y^{2} + 10 x^{2} y^{3} + 5 x y^{4} + y^{5} \end{array}$

The expansions of (x+y)2 ${(x + y)}^{2}$ and (x+y)3 ${(x + y)}^{3}$ should be familiar to you from Section P.3; the last two are left for you to verify (see Exercises 65 and 66 in Section 8.4). The key point is that each is a product of the previous binomial and (x+y). $(x + y) .$ For example, (x+y)4=(x+y)3(x+y). ${(x + y)}^{4} = {(x + y)}^{3} (x + y) .$

The patterns for expansions of (x+y)n ${(x + y)}^{n}$ (with n=1, 2, 3, 4, 5 $n = 1, 2, 3, 4, 5$ ) suggest the following:

The expansion of (x+y)n ${(x + y)}^{n}$ has n+1 $n + 1$ terms.
The sum of the exponents on x and y in each term equals n.
The exponent on x starts at n(xn=xn⋅y0) $n (x^{n} = x^{n} \cdot y^{0})$ in the first term and decreases by 1 for each term until it is 0 in the last term (x0⋅yn=yn). $(x^{0} \cdot y^{n} = y^{n}) .$
The exponent on y starts at 0 (xn=xn⋅y0) $0 (x^{n} = x^{n} \cdot y^{0})$ in the first term and increases by 1 for each term until it is n in the last term (x0⋅yn=yn). $(x^{0} \cdot y^{n} = y^{n}) .$
The variables x and y have symmetric roles. That is, replacing x with y and y with x in the expansion of (x+y)n ${(x + y)}^{n}$ yields the same terms, just in the reverse order.

You may also have noticed that the coefficients of the first and last terms are both 1 and the coefficients of the second and the next-to-last terms are equal. In general, the coefficients of

x n - j y j and x j y n - j

$x^{n - j} y^{j} and x^{j} y^{n - j}$

are equal for j=0, 1, 2, …, n. $j = 0, 1, 2, \dots, n .$

The coefficients in the expansion of (x+y)n ${(x + y)}^{n}$ are called the binomial coefficients.

Pascal’s Triangle

1 Use Pascal’s Triangle to compute binomial coefficients.

As early as a.d. 1100, the Chinese scholar Chia Hsien had discovered the secret of the binomial coefficients that was later rediscovered by the French philosopher and mathematician Blaise Pascal (1623–1662). To understand Chia Hsien’s and Pascal’s construction of binomial coefficients, let’s first look at the coefficients in these binomial expansions:

(x + y) 0 (x + y) 1 (x + y) 2 (x + y) 3 (x + y) 4 (x + y) 5 = = = = = = 1 1 x + 1 y 1 x 2 + 2 x y + 1 y 2 1 x 3 + 3 x 2 y + 3 x y 2 + 1 y 3 1 x 4 + 4 x 3 y + 6 x 2 y 2 + 4 x y 3 + 1 y 4 1 x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 x y 4 + 1 y 5

$\begin{array}{rcc} {(x + y)}^{0} & = & 1 \\ {(x + y)}^{1} & = & 1 x + 1 y \\ {(x + y)}^{2} & = & 1 x^{2} + 2 x y + 1 y^{2} \\ {(x + y)}^{3} & = & 1 x^{3} + 3 x^{2} y + 3 x y^{2} + 1 y^{3} \\ {(x + y)}^{4} & = & 1 x^{4} + 4 x^{3} y + 6 x^{2} y^{2} + 4 x y^{3} + 1 y^{4} \\ {(x + y)}^{5} & = & 1 x^{5} + 5 x^{4} y + 10 x^{3} y^{2} + 10 x^{2} y^{3} + 5 x y^{4} + 1 y^{5} \end{array}$

If we remove the variables and the plus signs and list only the coefficients, we get a triangle of numbers composed of the binomial coefficients. This triangle of numbers is known as Pascal’s Triangle.

Note the symmetry in Pascal’s Triangle. If the triangle were folded vertically down the middle, the numbers on each side of the crease would match. To create a new bottom row in the triangle, put the number 1 in the first and last places of the new row and add two neighboring entries in the previous row.

The top row is called the zeroth row because it corresponds to the binomial expansion of (x+y)0. ${(x + y)}^{0} .$ The next row is called the first row because it corresponds to the binomial expansion of (x+y)1. ${(x + y)}^{1} .$ All of the rows are named so that the nth row corresponds to the coefficients of (x+y)n. ${(x + y)}^{n} .$

2 Use Pascal’s Triangle to expand a binomial power.

Example 1 Using Pascal’s Triangle to Expand a Binomial Power

Expand (4y−2x)5. ${(4 y - 2 x)}^{5} .$

Solution

From the fifth row of Pascal’s Triangle, we see that the binomial coefficients are

1, 5, 10, 10, 5, 1 .

$1, 5, 10, 10, 5, 1 .$

We must make some changes in the expansion

(x + y) 5 = x 5 + 5 x 4 y + 10 x 3 y 2 + 10 x 2 y 3 + 5 x y 4 + y 5

${(x + y)}^{5} = x^{5} + 5 x^{4} y + 10 x^{3} y^{2} + 10 x^{2} y^{3} + 5 x y^{4} + y^{5}$

to get the expansion for (4y−2x)5: ${(4 y - 2 x)}^{5} :$

Replace x with 4y.
Replace y with −2x. $- 2 x .$

$(4 y - 2 x) 5 = = [4 y + (- 2 x)] 5 = (4 y) 5 + 5 (4 y) 4 (- 2 x) + 10 (4 y) 3 (- 2 x) 2 + 10 (4 y) 2 (- 2 x) 3 + 5 (4 y) (- 2 x) 4 + (- 2 x) 5 1024 y 5 - 2560 y 4 x + 2560 y 3 x 2 - 1280 y 2 x 3 + 320 y x 4 - 32 x 5$ $\begin{array}{rcl} {(4 y - 2 x)}^{5} & = & {[4 y + (- 2 x)]}^{5} = {(4 y)}^{5} + 5 {(4 y)}^{4} (- 2 x) \\ + 10 {(4 y)}^{3} {(- 2 x)}^{2} + 10 {(4 y)}^{2} {(- 2 x)}^{3} + 5 (4 y) {(- 2 x)}^{4} + {(- 2 x)}^{5} \\ = & 1024 y^{5} - 2560 y^{4} x + 2560 y^{3} x^{2} - 1280 y^{2} x^{3} + 320 y x^{4} - 32 x^{5} \end{array}$

We note that expanding a difference results in alternating signs between terms.

Practice Problem 1

Expand (3y−x)6. ${(3 y - x)}^{6} .$

The Binomial Theorem

3 Use the Binomial Theorem to expand a binomial power.

The coefficients in a binomial expansion can be computed by using ratios of certain factorials.

We first introduce the symbol (nr) $(\begin{array}{l} n \\ r \end{array})$ .

The symbol (nr) $(\begin{array}{l} n \\ r \end{array})$ is read “n choose r.” It can be shown that (nr) $(\begin{array}{l} n \\ r \end{array})$ is the number of ways of choosing a subset containing exactly r elements from a set with n elements.

Example 2 Evaluating (nr) $(\begin{array}{l} n \\ r \end{array})$

Evaluate each expression.

(41) $(\begin{array}{l} 4 \\ 1 \end{array})$
(53) $(\begin{array}{l} 5 \\ 3 \end{array})$
(90) $(\begin{array}{l} 9 \\ 0 \end{array})$
(3535) $(\begin{array}{l} 35 \\ 35 \end{array})$

Solution

(41)=4!1!(4−1)!=4!1! 3!=4⋅3⋅2⋅11(3⋅2⋅1)=41=4 $(\begin{array}{l} 4 \\ 1 \end{array}) = \frac{4!}{1! (4 - 1)!} = \frac{4!}{1! 3!} = \frac{4 \cdot 3 \cdot 2 \cdot 1}{1 (3 \cdot 2 \cdot 1)} = \frac{4}{1} = 4$
(53)=5!3!(5−3!)=5!3! 2!=5⋅4⋅3!3! 2!=5⋅42=5⋅2=10 $(\begin{array}{l} 5 \\ 3 \end{array}) = \frac{5!}{3! (5 - 3!)} = \frac{5!}{3! 2!} = \frac{5 \cdot 4 \cdot 3!}{3! 2!} = \frac{5 \cdot 4}{2} = 5 \cdot 2 = 10$
(90)=9!0!(9−0)!=9!0! 9!=11=1Recall that 0!=1. $\begin{matrix} (\begin{array}{l} 9 \\ 0 \end{array}) = \frac{9!}{0! (9 - 0)!} = \frac{9!}{0! 9!} = \frac{1}{1} = 1 & Recall that 0! = 1 . \end{matrix}$
(3535)=35!35!(35−35)!=35!35! 0!=11=1 $(\begin{array}{l} 35 \\ 35 \end{array}) = \frac{35!}{35! (35 - 35)!} = \frac{35!}{35! 0!} = \frac{1}{1} = 1$

Practice Problem 2

Evaluate each expression.
1. (62) $(\begin{array}{l} 6 \\ 2 \end{array})$
2. (129) $(\begin{array}{c} 12 \\ 9 \end{array})$

The numbers (nr) $(\begin{array}{l} n \\ r \end{array})$ show up as the coefficients in the expansion of a binomial power. For example, (x+y)4 ${(x + y)}^{4}$ can be written as either

(x + y) 4 = x 4 + 4 x 3 y + 6 x 2 y 2 + 4 x y 3 + y 4

$\begin{array}{lcccc} {(x + y)}^{4} = & x^{4} + & 4 x^{3} y + & 6 x^{2} y^{2} + & 4 x y^{3} + y^{4} \end{array}$

(x + y) 4 = (40) x 4 + (41) x 3 y + (42) x 2 y 2 + (43) x y 3 + (44) y 4 .

${(x + y)}^{4} =^{} (\begin{array}{l} 4 \\ 0 \end{array}) x^{4} +^{} (\begin{array}{l} 4 \\ 1 \end{array}) x^{3} y + (\begin{array}{l} 4 \\ 2 \end{array}) x^{2} y^{2} +^{} (\begin{array}{l} 4 \\ 3 \end{array}) x y^{3} + (\begin{array}{l} 4 \\ 4 \end{array}) y^{4} .$

This result is the Binomial Theorem (for n=4 $n = 4$ ), which can be proved by mathematical induction. It provides an efficient method for expanding a binomial power. (See Exercise 86.) The Binomial Theorem can be used to expand a binomial power directly, without reference to Pascal’s Triangle. This technique is particularly useful in expanding large powers of a binomial. For example, the expansion of (x+y)20 ${(x + y)}^{20}$ would require that you produce 20 rows of the Pascal Triangle.

Side Note

The kth term in the expansion of (x+y)n ${(x + y)}^{n}$ has coefficient (nk−1) $(\begin{array}{c} n \\ k - 1 \end{array})$ .

Example 3 Expanding a Binomial Power by Using the Binomial Theorem

Find the binomial expansion of (x−3y)4. ${(x - 3 y)}^{4} .$

Solution

Replace y with −3y $- 3 y$ in the expansion of (x+y)4 ${(x + y)}^{4}$

(x - 3 y) 4 = = = = = = [x + (- 3 y)] 4 (40) x 4 + (41) x 3 (- 3 y) + (42) x 2 (- 3 y) 2 + (43) x (- 3 y) 3 + (44) (- 3 y) 4 4 ! 0 ! 4 ! x 4 + 4 ! 1 ! 3 ! x 3 (- 3 y) + 4 ! 2 ! 2 ! x 2 (- 3 y) 2 + 4 ! 3 ! 1 ! x (- 3 y) 3 + 4 ! 4 ! 0 ! (- 3 y) 4 x 4 + 4 \cdot 3 ! 1 ! 3 ! x 3 (- 3 y) + 4 \cdot 3 \cdot 2 ! 2 ! 2 ! x 2 (- 3 y) 2 + 4 \cdot 3 ! 3 ! 1 ! x (- 3 y) 3 + (- 3 y) 4 x 4 + 4 x 3 (- 3 y) + 4 \cdot 3 2 ! x 2 (9 y 2) + 4 x (- 27 y 3) + (81 y 4) x 4 - 12 x 3 y + 54 x 2 y 2 - 108 x y 3 + 81 y 4

$\begin{array}{rcl} {(x - 3 y)}^{4} & = & {[x + (- 3 y)]}^{4} \\ = & (\begin{array}{l} 4 \\ 0 \end{array}) x^{4} + (\begin{array}{l} 4 \\ 1 \end{array}) x^{3} (- 3 y) + (\begin{array}{l} 4 \\ 2 \end{array}) x^{2} {(- 3 y)}^{2} + (\begin{array}{l} 4 \\ 3 \end{array}) x {(- 3 y)}^{3} + (\begin{array}{l} 4 \\ 4 \end{array}) {(- 3 y)}^{4} \\ = & \frac{4!}{0! 4!} x^{4} + \frac{4!}{1! 3!} x^{3} (- 3 y) + \frac{4!}{2! 2!} x^{2} {(- 3 y)}^{2} + \frac{4!}{3! 1!} x {(- 3 y)}^{3} + \frac{4!}{4! 0!} {(- 3 y)}^{4} \\ = & x^{4} + \frac{4 \cdot 3!}{1! 3!} x^{3} (- 3 y) + \frac{4 \cdot 3 \cdot 2!}{2! 2!} x^{2} {(- 3 y)}^{2} + \frac{4 \cdot 3!}{3! 1!} x {(- 3 y)}^{3} + {(- 3 y)}^{4} \\ = & x^{4} + 4 x^{3} (- 3 y) + \frac{4 \cdot 3}{2!} x^{2} (9 y^{2}) + 4 x (- 27 y^{3}) + (81 y^{4}) \\ = & x^{4} - 12 x^{3} y + 54 x^{2} y^{2} - 108 x y^{3} + 81 y^{4} \end{array}$

Practice Problem 3

Find the binomial expansion of (3x−y)4. ${(3 x - y)}^{4} .$

Binomial Coefficients

4 Find the coefficient of a term in a binomial expansion.

The Binomial coefficients (nr) $(\begin{array}{l} n \\ r \end{array})$ are useful for finding a particular coefficient or term in a binomial expansion.

Example 4 Finding a Particular Coefficient in a Binomial Expansion

Find the coefficient of x9y3 $x^{9} y^{3}$ in the expansion of (x+y)12. ${(x + y)}^{12} .$

Solution

The coefficient of xn−ryr is (nr) $x^{n - r} y^{r} is (\begin{array}{l} n \\ r \end{array})$ . Here n=12, r=3 $n = 12, r = 3$ , and n−r=9 $n - r = 9$ , so the coefficient of x9y3 $x^{9} y^{3}$ is

(n r) = (123) = 12 ! 3 ! ( 12 - 3 ! ) = 12 ! 3 ! 9 ! = 12 \cdot 11 \cdot 10 \cdot 9 ! 3 ! 9! = 220 .

$(\begin{array}{l} n \\ r \end{array}) = (\begin{array}{c} 12 \\ 3 \end{array}) = \frac{12!}{3! (12 - 3!)} = \frac{12!}{3! 9!} = \frac{12 \cdot 11 \cdot 10 \cdot 9!}{3! 9!} = 220 .$

Practice Problem 4

Find the coefficient of x3y9 $x^{3} y^{9}$ in the expansion of (x+y)12. ${(x + y)}^{12} .$

The method illustrated in Example 4 allows us to find any particular term in a binomial expansion without writing out the complete expansion.

Example 5 Finding a Particular Term in a Binomial Expansion

Find the term containing x10 $x^{10}$ in the expansion of (x+2a)15. ${(x + 2 a)}^{15} .$

Solution

We begin with the formula for the term containing the factor xr. $x^{r} .$

(n n - r) x r y n - r = = = = = = (15 15 - 10) x 10 (2 a) 15 - 10 (155) x 10 (2 a) 5 15 ! 5 ! ( 15 - 5 ) ! x 10 25 a 5 15 ! 5 ! 10 ! \cdot 32 x 10 a 5 15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10 ! 5 ! \cdot 10 ! \cdot 32 x 10 a 5 96,096 x 10 a 5 Replace n with 15, r with 10, and y with 2 a . Simplify . (n r) = n ! r ! ( n - r ) !; (2 a) 5 = 25 a 5 25 = 32 Use a calculator .

$\begin{array}{rcll} (\begin{array}{c} n \\ n - r \end{array}) x^{r} y^{n - r} & = & (\begin{array}{c} 15 \\ 15 - 10 \end{array}) x^{10} {(2 a)}^{15 - 10} & \begin{array}{l} Replace n with 15, r with 10, \\ and y with 2 a . \end{array} \\ = & (\begin{array}{c} 15 \\ 5 \end{array}) x^{10} {(2 a)}^{5} & Simplify . \\ = & \frac{15!}{5! (15 - 5)!} x^{10} 2^{5} a^{5} & (\begin{array}{c} n \\ r \end{array}) = \frac{n!}{r! (n - r)!}; {(2 a)}^{5} = 2^{5} a^{5} \\ = & \frac{15!}{5! 10!} \cdot 32 x^{10} a^{5} & 2^{5} = 32 \\ = & \frac{15 \cdot 14 \cdot 13 \cdot 12 \cdot 11 \cdot 10!}{5! \cdot 10!} \cdot 32 x^{10} a^{5} & Use a calculator . \\ = & 96,096 x^{10} a^{5} \end{array}$

Practice Problem 5

Find the term containing x3 $x^{3}$ in the expansion of (x+2a)15. ${(x + 2 a)}^{15} .$

Recall that, assuming decreasing powers of x, the kth term in the expansion of (x+y)n ${(x + y)}^{n}$ is (nk−1)xn−k+1yk−1. $(\begin{array}{c} n \\ k - 1 \end{array}) x^{n - k + 1} y^{k - 1} .$

Example 6 Finding a Specific Term in the Expansion of (x+y)n ${(x + y)}^{n}$

Find the 15th term in the expansion of (2x−1)18 ${(2 x - 1)}^{18}$ . Assume decreasing powers of x.

Solution

The kth term in the expansion of (x+y)n ${(x + y)}^{n}$ is (nk−1)xn−k+1yk−1. $(\begin{array}{c} n \\ k - 1 \end{array}) x^{n - k + 1} y^{k - 1} .$ Note that (2x−1)18=(2x+(−1))18. ${(2 x - 1)}^{18} = {(2 x + (- 1))}^{18} .$ If we replace x with 2x and y with −1, $- 1,$ then n with 18, and k with 15, we get:

(1814) x 18 - 15 + 1 y 15 - 1 (1814) (2 x) 18 - 14 (- 1) 14 (1814) (2 x) 18 - 14 (- 1) 14 = 18 ! 14 ! 4 ! 24 x 4 The 15 th term of (x + y) 18 Replace x with 2 x and y with - 1 in (1814) x 18 - 14 y 14 . (2 x) 18 - 14 = (2 x) 4 = 24 x 4; (- 1) 14 = 1

$\begin{array}{l} (\begin{array}{c} 18 \\ 14 \end{array}) x^{18 - 15 + 1} y^{15 - 1} & The 15^{th} term of {(x + y)}^{18} \\ (\begin{array}{c} 18 \\ 14 \end{array}) {(2 x)}^{18 - 14} {(- 1)}^{14} & Replace x with 2 x and y with - 1 in (\begin{array}{c} 18 \\ 14 \end{array}) x^{18 - 14} y^{14} . \\ (\begin{array}{c} 18 \\ 14 \end{array}) {(2 x)}^{18 - 14} {(- 1)}^{14} = \frac{18!}{14! 4!} 2^{4} x^{4} & {(2 x)}^{18 - 14} = {(2 x)}^{4} = 2^{4} x^{4}; {(- 1)}^{14} = 1 \end{array}$

The 15th term in the expansion of (2x−1)18 ${(2 x - 1)}^{18}$ is 18!14!4! 24x4=48960x4. Simplify $\frac{18!}{14! 4!} 2^{4} x^{4} = 48960 x^{4} . Simplify$

Practice Problem 6

Find the fourth term in the expansion of (x−3)12 ${(x - 3)}^{12}$ . Assume decreasing powers of x.

Section 8.5 Exercises

Concepts and Vocabulary

The expansion of (x+y)n ${(x + y)}^{n}$ has terms.
In the expansion of (x+y)5, ${(x + y)}^{5},$ the coefficient of x2y3 $x^{2} y^{3}$ is (nr) $(\begin{array}{l} n \\ r \end{array})$ when n=−−−−−−−−−−−−−−− $n = \underline{}$ and r=−−−−−−−−−−−−−−− $r = \underline{}$ .
Expanding a difference such as (2x−y)10 ${(2 x - y)}^{10}$ results in between terms.
For any positive integer n, (nn)=−−−−−−−−−−−−−−− $(\begin{array}{l} n \\ n \end{array}) = \underline{}$ .
True or False. Every coefficient in the expansion of (x+y)n, ${(x + y)}^{n},$ for n>1, $n > 1,$ appears exactly twice.
True or False. If n>3, $n > 3,$ then 2(n2)=n. $2 (\begin{array}{l} n \\ 2 \end{array}) = n .$
True or False. If n≥1 $n \geq 1$ , then (x−y)n ${(x - y)}^{n}$ has n+1 $n + 1$ terms.
True or False. The coefficient of the sixth term in the expansion of (x+y)55 ${(x + y)}^{55}$ is (555) $(\begin{array}{l} 55 \\ 5 \end{array})$ , assuming descending powers of x.

Building Skills

In Exercises 9–24, evaluate each expression.

6!3! $\frac{6!}{3!}$
11!9! $\frac{11!}{9!}$
12!11! $\frac{12!}{11!}$
3!0! $\frac{3!}{0!}$
(64) $(\begin{array}{l} 6 \\ 4 \end{array})$
(63) $(\begin{array}{l} 6 \\ 3 \end{array})$
(90) $(\begin{array}{l} 9 \\ 0 \end{array})$
(120) $(\begin{array}{c} 12 \\ 0 \end{array})$
(71) $(\begin{array}{l} 7 \\ 1 \end{array})$
(73) $(\begin{array}{l} 7 \\ 3 \end{array})$
(4545) $(\begin{array}{l} 45 \\ 45 \end{array})$
(450) $(\begin{array}{c} 45 \\ 0 \end{array})$
(10098) $(\begin{array}{c} 100 \\ 98 \end{array})$
(1002) $(\begin{array}{c} 100 \\ 2 \end{array})$
(2119) $(\begin{array}{c} 21 \\ 19 \end{array})$
(152) $(\begin{array}{c} 15 \\ 2 \end{array})$

In Exercises 25–32, use Pascal’s Triangle to expand each binomial.

(x+2)4 ${(x + 2)}^{4}$
(x+3)4 ${(x + 3)}^{4}$
(x−2)5 ${(x - 2)}^{5}$
(3−x)5 ${(3 - x)}^{5}$
(2−3x)3 ${(2 - 3 x)}^{3}$
(3−2x)3 ${(3 - 2 x)}^{3}$
(2x+3y)4 ${(2 x + 3 y)}^{4}$
(2x+5y)4 ${(2 x + 5 y)}^{4}$

In Exercises 33–54, use either the Binomial Theorem or Pascal’s Triangle to expand each binomial.

(x+1)4 ${(x + 1)}^{4}$
(x+2)4 ${(x + 2)}^{4}$
(x−1)5 ${(x - 1)}^{5}$
(1−x)5 ${(1 - x)}^{5}$
(y−3)3 ${(y - 3)}^{3}$
(2−y)5 ${(2 - y)}^{5}$
(x+y)6 ${(x + y)}^{6}$
(x−y)6 ${(x - y)}^{6}$
(1+3y)5 ${(1 + 3 y)}^{5}$
(2x+1)5 ${(2 x + 1)}^{5}$
(2x+1)4 ${(2 x + 1)}^{4}$
(3x−1)4 ${(3 x - 1)}^{4}$
(x−2y)3 ${(x - 2 y)}^{3}$
(2x−y)3 ${(2 x - y)}^{3}$
(2x+y)4 ${(2 x + y)}^{4}$
(3x−2y)4 ${(3 x - 2 y)}^{4}$
(x2+2)7 ${(\frac{x}{2} + 2)}^{7}$
(2−x2)7 ${(2 - \frac{x}{2})}^{7}$
(a2−13)4 ${(a^{2} - \frac{1}{3})}^{4}$
(12−a2)4 ${(\frac{1}{2} - a^{2})}^{4}$
(1x+y)3 ${(\frac{1}{x} + y)}^{3}$
(x+2y)3 ${(x + \frac{2}{y})}^{3}$

In Exercises 55–62, find the specified coefficient.

for x2y7 $x^{2} y^{7}$ in (x+y)9 ${(x + y)}^{9}$
for x10y5 $x^{10} y^{5}$ in (x−y)15 ${(x - y)}^{15}$
for x5 $x^{5}$ in (x+3)8 ${(x + 3)}^{8}$
for x7 $x^{7}$ in (x−2)10 ${(x - 2)}^{10}$
for x2y7 $x^{2} y^{7}$ in (2x+3y)9 ${(2 x + 3 y)}^{9}$
for x3y4 $x^{3} y^{4}$ in (x−2y)7 ${(x - 2 y)}^{7}$
for x10y12 $x^{10} y^{12}$ in (x2+y3)9 ${(x^{2} + y^{3})}^{9}$
for x18y6 $x^{18} y^{6}$ in (x3+2y2)9 ${(x^{3} + 2 y^{2})}^{9}$

In Exercises 63–76, find the specified term. In Exercises 71–76 assume descending powers of x.

(x+y)10; ${(x + y)}^{10};$ term containing x7 $x^{7}$
(x+y)10; ${(x + y)}^{10};$ term containing y7 $y^{7}$
(x−2)12; ${(x - 2)}^{12};$ term containing x3 $x^{3}$
(2−x)12; ${(2 - x)}^{12};$ term containing x3 $x^{3}$
(2x+3y)8; ${(2 x + 3 y)}^{8};$ term containing x6 $x^{6}$
(2x+3y)8; ${(2 x + 3 y)}^{8};$ term containing y6 $y^{6}$
(5x−2y)11; ${(5 x - 2 y)}^{11};$ term containing y9 $y^{9}$
(7x−y)15; ${(7 x - y)}^{15};$ term containing x9 $x^{9}$
(x−1)9 ${(x - 1)}^{9}$ ; seventh term.
(x−1)9 ${(x - 1)}^{9}$ ; fourth term.
(2x−y)7 ${(2 x - y)}^{7}$ ; fifth term.
(−2x+y)7 ${(- 2 x + y)}^{7}$ ; seventh term.
(x+3y)10 ${(x + 3 y)}^{10}$ ; third term.
(x+0.5y)8 ${(x + 0.5 y)}^{8}$ ; sixth term.
Find the value of (1.2)5 ${(1.2)}^{5}$ by writing it in the form (1+0.2)5 ${(1 + 0.2)}^{5}$ and applying the Binomial Theorem.
Use the method in Exercise 77 to evaluate each expression.
1. (2.9)4 ${(2.9)}^{4}$
2. (10.4)3 ${(10.4)}^{3}$

Beyond the Basics

Find the middle term in the expansion of (x−−√−2x2)10. ${(\sqrt{x} - \frac{2}{x^{2}})}^{10} .$
Find the middle term in the expansion of (x−−√+1x2)10. ${(\sqrt{x} + \frac{1}{x^{2}})}^{10} .$
Find the middle term in the expansion of (1−x2y−3)12. ${(1 - x^{2} y^{- 3})}^{12} .$
Show that the middle term in the expansion of (1+x)2n ${(1 + x)}^{2 n}$ is

$1 \cdot 3 \cdot 5 \cdot \dots \cdot ( 2 n - 1 ) n ! 2 n x n .$ $\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2 n - 1)}{n!} 2^{n} x^{n} .$
Prove that (n0)+(n1)+(n2)+⋯+(nn)=2n. $(\begin{array}{l} n \\ 0 \end{array}) + (\begin{array}{l} n \\ 1 \end{array}) + (\begin{array}{l} n \\ 2 \end{array}) + \dots + (\begin{array}{l} n \\ n \end{array}) = 2^{n} .$
Prove that (n0)−(n1)+(n2)−(n3)+⋯+(−1)n(nn)=0. $(\begin{array}{l} n \\ 0 \end{array}) - (\begin{array}{l} n \\ 1 \end{array}) + (\begin{array}{l} n \\ 2 \end{array}) - (\begin{array}{l} n \\ 3 \end{array}) + \dots + {(- 1)}^{n} (\begin{array}{l} n \\ n \end{array}) = 0.$
Prove that (kj)+(kj−1)=(k+1j). $(\begin{array}{c} k \\ j \end{array}) + (\begin{array}{c} k \\ j - 1 \end{array}) = (\begin{array}{c} k + 1 \\ j \end{array}) .$
Prove the Binomial Theorem by using the principle of mathematical induction.

[Hint:

$(x + y) k + 1 = = (x + y) (x + y) k = (x + y) \sum j = 0 k (k j) x k - j y j \sum j = 0 k (k j) x k + 1 - j y j + \sum j = 0 k (k j) x k - j y j + 1 .$ $\begin{array}{rcl} {(x + y)}^{k + 1} & = & (x + y) {(x + y)}^{k} = (x + y) \sum_{j = 0}^{k} (\begin{array}{l} k \\ j \end{array}) x^{k - j} y^{j} \\ = & \sum_{j = 0}^{k} (\begin{array}{l} k \\ j \end{array}) x^{k + 1 - j} y^{j} + \sum_{j = 0}^{k} (\begin{array}{l} k \\ j \end{array}) x^{k - j} y^{j + 1} . \end{array}$

Use Exercise 85 to show that

$(x + y) k + 1 = \sum j = 0 k + 1 (k + 1 j) x k + 1 - j y j .]$ ${(x + y)}^{k + 1} = \sum_{j = 0}^{k + 1} (\begin{array}{c} k + 1 \\ j \end{array}) x^{k + 1 - j} y^{j} .]$

In Exercises 87–89, find the value of the expression without expanding any term.

(2x−1)4+4(2x−1)3(3−2x)+6(2x−1)2(3−2x)2+4(2x−1)(3−2x)3+(3−2x)4 $\begin{array}{l} {(2 x - 1)}^{4} + 4 {(2 x - 1)}^{3} (3 - 2 x) + 6 {(2 x - 1)}^{2} {(3 - 2 x)}^{2} \\ + 4 (2 x - 1) {(3 - 2 x)}^{3} + {(3 - 2 x)}^{4} \end{array}$
(x+1)4−4(x+1)3(x−1)+6(x+1)2(x−1)2−4(x+1)(x−1)3+(x−1)4 $\begin{array}{l} {(x + 1)}^{4} - 4 {(x + 1)}^{3} (x - 1) + 6 {(x + 1)}^{2} {(x - 1)}^{2} \\ - 4 (x + 1) {(x - 1)}^{3} + {(x - 1)}^{4} \end{array}$
(3x−1)5+5(3x−1)4(1−2x)+10(3x−1)3(1−2x)2+10(3x−1)2(1−2x)3+5(3x−1)(1−2x)4+(1−2x)5 $\begin{array}{l} {(3 x - 1)}^{5} + 5 {(3 x - 1)}^{4} (1 - 2 x) \\ + 10 {(3 x - 1)}^{3} {(1 - 2 x)}^{2} +^{} 10 {(3 x - 1)}^{2} {(1 - 2 x)}^{3} \\ + 5 (3 x - 1) {(1 - 2 x)}^{4} + {(1 - 2 x)}^{5} \end{array}$
Find the constant term in each expansion.
1. (x2−1x)9 ${(x^{2} - \frac{1}{x})}^{9}$
2. (x−−√−2x2)10 ${(\sqrt{x} - \frac{2}{x^{2}})}^{10}$
If the constant term in the expansion of (kx−1x2)6 ${(k x - \frac{1}{x^{2}})}^{6}$ is 240, find k.
If the constant term in the expansion of (x3+kx8)11 ${(x^{3} + \frac{k}{x^{8}})}^{11}$ is 1320, find k.
Show that there is no constant term in the expansion of (2x2−14x)11. ${(2 x^{2} - \frac{1}{4 x})}^{11} .$

Critical Thinking / Discussion / Writing

Use the binomial expansion of ${(x + y)}^{6},$ with $x = 1$ and $y = 1,$ to show that

$2^{6} = (\begin{array}{l} 6 \\ 0 \end{array}) + (\begin{array}{l} 6 \\ 1 \end{array}) + (\begin{array}{l} 6 \\ 2 \end{array}) + (\begin{array}{l} 6 \\ 3 \end{array}) + (\begin{array}{l} 6 \\ 4 \end{array}) + (\begin{array}{l} 6 \\ 5 \end{array}) + (\begin{array}{l} 6 \\ 6 \end{array}) .$
Use the binomial expansion of ${(x + y)}^{10}$ to show that

$\begin{array}{l} (\begin{array}{c} 10 \\ 0 \end{array}) - (\begin{array}{c} 10 \\ 1 \end{array}) + (\begin{array}{c} 10 \\ 2 \end{array}) - (\begin{array}{c} 10 \\ 3 \end{array}) + (\begin{array}{c} 10 \\ 4 \end{array}) - (\begin{array}{c} 10 \\ 5 \end{array}) \\ + (\begin{array}{c} 10 \\ 6 \end{array}) - (\begin{array}{c} 10 \\ 7 \end{array}) + (\begin{array}{c} 10 \\ 8 \end{array}) - (\begin{array}{c} 10 \\ 9 \end{array}) + (\begin{array}{c} 10 \\ 10 \end{array}) = 0. \end{array}$
Use the binomial expansion of ${(x + y)}^{2}$ to show that if $x > 0$ and $y > 0,$ then ${(x + y)}^{2} > x^{2} + y^{2} .$
Use the binomial expansion of ${(x + y)}^{n}$ to show that if $x > 0$ and $y > 0,$ then ${(x + y)}^{n} > x^{n} + y^{n} .$
Use the binomial expansion of ${(1 + x)}^{n}$ to show that if $x > 0,$ then ${(1 + x)}^{n} > 1 + n x .$

Getting Ready for the Next Section

In Exercises 99–106, write each expression using factorial notation.

$5 \cdot 6 \cdot 7 \cdot 8$
$10 \cdot 9 \cdot 8 \cdot 7 \cdot 6$
$2 \cdot 4 \cdot 6 \cdot 8 \cdot 10 \cdot 12$
$3 \cdot 6 \cdot 9 \cdot 12 \cdot 15$
Compute $\frac{12!}{10!} .$
Compute $\frac{n!}{(n - r)!}$ for $n = 100$ and $r = 2 .$
Compute $\frac{8!}{5! 3!} .$
Compute $\frac{n!}{(n - r)! r!}$ for $n = 10$ and $r = 3 .$

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Table of Contents for Section 8.5 The Binomial Theorem

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Table of Contents for
Section 8.5 The Binomial Theorem