MX Precalc Binomial Theorem

Section 11.6Binomial Theorem

Corequisite Skills review (optional warm-up)

Learning Objectives

  • Use Pascal’s Triangle to expand a binomial. (IA 12.4.1)

Objective 1: Use Pascal’s Triangle to expand a binomial. (IA 12.4.1)

Pascal’s triangle helps us find the coefficients of the terms in the expansion of a binomial.

To find the coefficients of the terms, we write our expansion again focusing on the coefficients. We rewrite the coefficients to the right forming an array of coefficients. The array to the right is called Pascal’s Triangle.

A plus b to the power of 0 equals 1. The top level of Pascal’s Triangle is 1. A plus b to the power of 1 equals 1 a plus 1 b. The second level of Pascal’s Triangle is 1, 1. A plus b to the power of 2 equals 1 a to the power of 2 plus 2 a b plus 1 b to the power of 2. The third level of Pascal’s Triangle is 1, 2, 1. A plus b to the power of 3 equals 1 a to the power of 3 plus 3 a to the power of 2 b plus 3 a b to the power of 2 plus 1 b to the power of 3. The fourth level of Pascal’s Triangle is 1,3,3,1. A plus b to the power of 4 equals 1 a to the power of 4 plus 4 a to the power of 3 b plus 6 a to the power of 2 b to the power of 2 plus 4 a b to the power of 3 plus 1 b to the power of 4. The fifth level of Pascal’s Triangle is 1, 4, 6, 4, 1. A plus b to the power of 5 equals 1 a to the power of 5 plus 5 a to the power of 4 b plus 10 a to the power of 3 b to the power of 2 plus 10 a to the power of 2 b to the power of 3. The sixth row of the Pascal’s Triangle is 1, 5, 10, 10, 5, 1.

Notice that in each expansion the powers of a in each term decrease from n to 0, and the powers of b increase from 0 to n.

Notice each number in the array is the sum of the two closest numbers in the row above. We can find the next row by starting and ending with one and then adding two adjacent numbers.

To find the coefficients of the expansion of the binomial \({(a+b)}^{n}\) , go to the row that has the value n as a second entry.

This figure shows Pascal’s Triangle. The first level is 1. The second level is 1, 1. The third level is 1, 2, 1. The fourth level is 1, 3, 3, 1. The fifth level is 1, 4, 6, 4, 1. The sixth level is 1, 5, 10, 10, 5, 1. The seventh level is 1, 6, 15, 20, 15, 6, 1.
Warm-up Example 1

Use Pascal’s Triangle to expand \({(x+y)}^{6}\) .

Table 1
Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 6.The image displays Pascal's Triangle, a triangular array of binomial coefficients. The first seven rows are shown, with the numbers in the last row (1, 6, 15, 20, 15, 6, 1) highlighted in red. Each number in the triangle is the sum of the two numbers directly above it, and the rows represent the coefficients of binomial expansions.
Write the expansion with the coefficients.The image displays the binomial expansion of (x+y)^6, with the numerical coefficients (1, 6, 15, 20, 15, 6, 1) filled in from Pascal's triangle. Underscores indicate the missing variable terms for each part of the expansion. The full expansion should be x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6.
Fill in the variable with the power of x decreasing from 6 to 0, and the power of y increasing from 0 to 6.The image displays the binomial expansion of (x+y)^6, which equals 1x^6 + 6x^5y^1 + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6x^1y^5 + 1y^6. The coefficients of the expansion are highlighted in red and underlined.
Binomial expansion of \({(x+y)}^{6}\) .The image shows the binomial expansion of (x+y) raised to the power of 6, which equals x^6 + 6x^5y^1 + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6x^1y^5 + y^6.
Warm-up Example 2

Use Pascal’s Triangle to expand \({(x+3)}^{5}\) .

Table 2

Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 5.Two binomial expressions: the general form (a+b)^n in red, and a specific application (x+3)^5 in black, demonstrating the binomial expansion concept. 		An illustration of Pascal's Triangle, a triangular array of binomial coefficients, with the fifth row (1 5 10 10 5 1) highlighted in red. Each number is the sum of the two directly above it.
Write the expansion with the coefficients.A mathematical expression displays the binomial expansion of (x+3) to the power of 5, showing the coefficients 1, 5, 10, 10, 5, and 1, with blank spaces for the variable terms.
Fill in the variable with the power of x decreasing from 5 to 0, and the power of 3 increasing from 0 to 5.
The binomial expansion of (x+3)⁵, showing the sum of terms where each term consists of a binomial coefficient (highlighted in red), a decreasing power of x, and an increasing power of 3. The coefficients are 1, 5, 10, 10, 5, 1, corresponding to Pascal's triangle for n=5. A mathematical equation illustrating the binomial expansion of (x + 3)^5. The expansion shows coefficients (1, 5, 10, 10, 5, 1) in red, multiplied by decreasing powers of 'x' and increasing powers of '3'.
Binomial expansion of \({(x+3)}^{5}\) .The image shows the mathematical equation representing the binomial expansion of (x+3) to the power of 5, which is equal to x^5 + 15x^4 + 90x^3 + 270x^2 + 405x^1 + 243.
Warm-up Example 3

Use Pascal’s Triangle to expand \({(3x-2)}^{4}\) .

Table 3

Go to Pascal’s Triangle and read off the coefficients from the row whose second entry is 4.The image displays two mathematical expressions: the general form of a binomial expansion, (a + b)^n, and a specific example, (3x - 2)^4. 		Pascal's triangle displaying binomial coefficients, where each number is the sum of the two directly above it. The fifth row (1, 4, 6, 4, 1) is highlighted.
Write the expansion with the coefficients.An algebraic expression showing the partial binomial expansion of (3x - 2)^4, with the coefficients 1, 4, 6, 4, 1 from Pascal's triangle highlighted in red.
Fill in the variable with the power of (3x) decreasing from 4 to 0, and the power of (-2) increasing from 0 to 4.

The image shows the mathematical identity (3x - 2) raised to the power of 4, which is expressed as being equal to the sum of (3x) and (-2), all raised to the power of 4. This demonstrates rewriting a subtraction within parentheses as an addition of a negative number. The image shows the binomial expansion of the expression (3x - 2)^4. It illustrates the application of the binomial theorem using the coefficients from Pascal's triangle (1, 4, 6, 4, 1) and demonstrating the decreasing powers of (3x) and increasing powers of (-2) for each term. The image displays the binomial expansion of (3x - 2)^4. It shows the application of the binomial theorem with coefficients 1, 4, 6, 4, 1 (in red) multiplied by the corresponding terms.
Binomial expansion of \({(3x-2)}^{4}\) .The image displays the binomial expansion of (3x - 2) raised to the power of 4, showing it equals 81x^4 - 216x^3 + 216x^2 - 96x + 16.

Practice Makes Perfect

Use Pascal’s Triangle to expand a binomial.

P1

Use Pascal’s Triangle to expand \({(a+b)}^{4}\) .

P2

Use Pascal’s Triangle to expand \({(y+3)}^{5}\) .

P3

Use Pascal’s Triangle to expand \({(2x-5)}^{3}\) .

A polynomial with two terms is called a binomial. We have already learned to multiply binomials and to raise binomials to powers, but raising a binomial to a high power can be tedious and time-consuming. In this section, we will discuss a shortcut that will allow us to find \({(x+y)}^{n}\) without multiplying the binomial by itself \(n\) times.

Identifying Binomial Coefficients

In Counting Principles, we studied combinations. In the shortcut to finding \({(x+y)}^{n},\) we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. In this case, we use the notation \((\begin{array}{l}n \\ r\end{array})\) instead of \(C(n,r),\) but it can be calculated in the same way. So

\[(\begin{array}{l}n \\ r\end{array})=C(n,r)=\frac{n!}{r!(n-r)!}\]

The combination \((\begin{array}{l}n \\ r\end{array})\) is called a binomial coefficient. An example of a binomial coefficient is \((\begin{array}{l}5 \\ 2\end{array})=C(5,2)=10.\)

Binomial Coefficients

If \(n\) and \(r\) are integers greater than or equal to 0 with \(n\ge r,\) then the binomial coefficient is

\[(\begin{array}{l}n \\ r\end{array})=C(n,r)=\frac{n!}{r!(n-r)!}\]

Q&A

Is a binomial coefficient always a whole number?

Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number.

Example 1

Find each binomial coefficient.

  • ⓐ \((\begin{array}{l}5 \\ 3\end{array})\)
  • ⓑ \((\begin{array}{l}9 \\ 2\end{array})\)
  • ⓒ \((\begin{array}{l}9 \\ 7\end{array})\)

Use the formula \(\binom{n}{r}=\frac{n!}{r!(n-r)!}\) for each pair of values.

Use the formula to calculate each binomial coefficient. You can also use the \(n{C}_{r}\) function on your calculator.

\[(\begin{array}{l}n \\ r\end{array})=C(n,r)=\frac{n!}{r!(n-r)!}\]

  • ⓐ \((\begin{array}{l}5 \\ 3\end{array})=\frac{5!}{3!(5-3)!}=\frac{5\cdot 4\cdot 3!}{3!2!}=10\)
  • ⓑ \((\begin{array}{l}9 \\ 2\end{array})=\frac{9!}{2!(9-2)!}=\frac{9\cdot 8\cdot 7!}{2!7!}=36\)
  • ⓒ \((\begin{array}{l}9 \\ 7\end{array})=\frac{9!}{7!(9-7)!}=\frac{9\cdot 8\cdot 7!}{7!2!}=36\)
Analysis

Notice that we obtained the same result for parts (b) and (c). If you look closely at the solution for these two parts, you will see that you end up with the same two factorials in the denominator, but the order is reversed, just as with combinations.

\[(\begin{array}{l}n \\ r\end{array})=(\begin{array}{l}n \\ n-r\end{array})\]

Find each binomial coefficient.

Try It #1
  • ⓐ \((\begin{array}{l}7 \\ 3\end{array})\)
  • ⓑ \((\begin{array}{l}11 \\ 4\end{array})\)
  • ⓐ35
  • ⓑ330
Did you get it?

Using the Binomial Theorem

When we expand \({(x+y)}^{n}\) by multiplying, the result is called a binomial expansion, and it includes binomial coefficients. If we wanted to expand \({(x+y)}^{52},\) we might multiply \((x+y)\) by itself fifty-two times. This could take hours! If we examine some simple binomial expansions, we can find patterns that will lead us to a shortcut for finding more complicated binomial expansions.

\[\begin{array}{l}{(x+y)}^{2}={x}^{2}+2xy+{y}^{2} \\ {(x+y)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\ {(x+y)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\end{array}\]

First, let’s examine the exponents. With each successive term, the exponent for \(x\) decreases and the exponent for \(y\) increases. The sum of the two exponents is \(n\) for each term.

Next, let’s examine the coefficients. Notice that the coefficients increase and then decrease in a symmetrical pattern. The coefficients follow a pattern:

\[(\begin{array}{l}n \\ 0\end{array}),(\begin{array}{l}n \\ 1\end{array}),(\begin{array}{l}n \\ 2\end{array}),...,(\begin{array}{l}n \\ n\end{array}).\]

These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.

\[\begin{array}{ll}{(x+y)}^{n} & =∑k=0n(\begin{array}{l}n \\ k\end{array}){x}^{n-k}{y}^{k} \\ & ={x}^{n}+(\begin{array}{l}n \\ 1\end{array}){x}^{n-1}y+(\begin{array}{l}n \\ 2\end{array}){x}^{n-2}{y}^{2}+...+(\begin{array}{l}n \\ n-1\end{array})x{y}^{n-1}+{y}^{n}\end{array}\]

Another way to see the coefficients is to examine the expansion of a binomial in general form, \(x+y,\) to successive powers 1, 2, 3, and 4.

\[\begin{array}{l}{(x+y)}^{1}=x+y \\ {(x+y)}^{2}={x}^{2}+2xy+{y}^{2} \\ {(x+y)}^{3}={x}^{3}+3{x}^{2}y+3x{y}^{2}+{y}^{3} \\ {(x+y)}^{4}={x}^{4}+4{x}^{3}y+6{x}^{2}{y}^{2}+4x{y}^{3}+{y}^{4}\end{array}\]

Can you guess the next expansion for the binomial \({(x+y)}^{5}?\)

Graph of the function f_2.
Figure 1

See Figure 1, which illustrates the following:

To determine the expansion on \({(x+y)}^{5},\) we see \(n=5,\) thus, there will be 5+1 = 6 terms. Each term has a combined degree of 5. In descending order for powers of \(x,\) the pattern is as follows:

The next expansion would be

\[{(x+y)}^{5}={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}.\]

But where do those coefficients come from? The binomial coefficients are symmetric. We can see these coefficients in an array known as Pascal's Triangle, shown in Figure 2. Pascal didn't invent the triangle. The underlying principles had been developed and written about for over 1500 years, first by the Indian mathematician (and poet) Pingala in the second century BCE. Others throughout Asia and Europe worked with the concepts throughout, and the triangle was first published in its graphical form by Omar Khayyam, an Iranian mathematician and astronomer, for whom the triangle is named in Iran. French mathematician Blaise Pascal repopularized it when he republished it and used it to solve a number of probability problems.

Pascal's Triangle
Figure 2

To generate Pascal’s Triangle, we start by writing a 1. In the row below, row 2, we write two 1’s. In the 3rd row, flank the ends of the rows with 1’s, and add \(1+1\) to find the middle number, 2. In the \(n\text{th}\) row, flank the ends of the row with 1’s. Each element in the triangle is the sum of the two elements immediately above it.

To see the connection between Pascal’s Triangle and binomial coefficients, let us revisit the expansion of the binomials in general form.

Pascal's Triangle expanded to show the values of the triangle as x and y terms with exponents
The Binomial Theorem

The Binomial Theorem is a formula that can be used to expand any binomial.

\[\begin{array}{ll}{(x+y)}^{n} & =∑k=0n(\begin{array}{l}n \\ k\end{array}){x}^{n-k}{y}^{k} \\ & ={x}^{n}+(\begin{array}{l}n \\ 1\end{array}){x}^{n-1}y+(\begin{array}{l}n \\ 2\end{array}){x}^{n-2}{y}^{2}+...+(\begin{array}{l}n \\ n-1\end{array})x{y}^{n-1}+{y}^{n}\end{array}\]

How To

Given a binomial, write it in expanded form.

  • Determine the value of \(n\) according to the exponent.
  • Evaluate the \(k=0\) through \(k=n\) using the Binomial Theorem formula.
  • Simplify.
Example 2

Write in expanded form.

  • ⓐ \({(x+y)}^{5}\)
  • ⓑ \({(3x-y)}^{4}\)

Use the Binomial Theorem to write out each term with its binomial coefficient, decreasing powers of the first term and increasing powers of the second.

  • ⓐSubstitute \(n=5\) into the formula. Evaluate the \(k=0\) through \(k=5\) terms. Simplify.

    \[\begin{array}{ll}{(x+y)}^{5} & =(\begin{array}{l}5 \\ 0\end{array}){x}^{5}{y}^{0}+(\begin{array}{l}5 \\ 1\end{array}){x}^{4}{y}^{1}+(\begin{array}{l}5 \\ 2\end{array}){x}^{3}{y}^{2}+(\begin{array}{l}5 \\ 3\end{array}){x}^{2}{y}^{3}+(\begin{array}{l}5 \\ 4\end{array}){x}^{1}{y}^{4}+(\begin{array}{l}5 \\ 5\end{array}){x}^{0}{y}^{5} \\ {(x+y)}^{5} & ={x}^{5}+5{x}^{4}y+10{x}^{3}{y}^{2}+10{x}^{2}{y}^{3}+5x{y}^{4}+{y}^{5}\end{array}\]

  • ⓑSubstitute \(n=4\) into the formula. Evaluate the \(k=0\) through \(k=4\) terms. Notice that \(3x\) is in the place that was occupied by \(x\) and that \(-y\) is in the place that was occupied by \(y.\) So we substitute them. Simplify.

    \[\begin{array}{ll}{(3x-y)}^{4} & =(\begin{array}{l}4 \\ 0\end{array}){(3x)}^{4}{(-y)}^{0}+(\begin{array}{l}4 \\ 1\end{array}){(3x)}^{3}{(-y)}^{1}+(\begin{array}{l}4 \\ 2\end{array}){(3x)}^{2}{(-y)}^{2}+(\begin{array}{l}4 \\ 3\end{array}){(3x)}^{1}{(-y)}^{3}+(\begin{array}{l}4 \\ 4\end{array}){(3x)}^{0}{(-y)}^{4} \\ {(3x-y)}^{4} & =81{x}^{4}-108{x}^{3}y+54{x}^{2}{y}^{2}-12x{y}^{3}+{y}^{4}\end{array}\]

Analysis

Notice the alternating signs in part b. This happens because \((-y)\) raised to odd powers is negative, but \((-y)\) raised to even powers is positive. This will occur whenever the binomial contains a subtraction sign.

Try It #2

Write in expanded form.

  • ⓐ \({(x-y)}^{5}\)
  • ⓑ \({(2x+5y)}^{3}\)
  • ⓐ \({x}^{5}-5{x}^{4}y+10{x}^{3}{y}^{2}-10{x}^{2}{y}^{3}+5x{y}^{4}-{y}^{5}\)
  • ⓑ \(8{x}^{3}+60{x}^{2}y+150x{y}^{2}+125{y}^{3}\)
Did you get it?

Using the Binomial Theorem to Find a Single Term

Expanding a binomial with a high exponent such as \({(x+2y)}^{16}\) can be a lengthy process.

Sometimes we are interested only in a certain term of a binomial expansion. We do not need to fully expand a binomial to find a single specific term.

Note the pattern of coefficients in the expansion of \({(x+y)}^{5}.\)

\[{(x+y)}^{5}={x}^{5}+(\begin{array}{l}5 \\ 1\end{array}){x}^{4}y+(\begin{array}{l}5 \\ 2\end{array}){x}^{3}{y}^{2}+(\begin{array}{l}5 \\ 3\end{array}){x}^{2}{y}^{3}+(\begin{array}{l}5 \\ 4\end{array})x{y}^{4}+{y}^{5}\]

The second term is \((\begin{array}{l}5 \\ 1\end{array}){x}^{4}y.\) The third term is \((\begin{array}{l}5 \\ 2\end{array}){x}^{3}{y}^{2}.\) We can generalize this result.

\[(\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\]

The (r+1)th Term of a Binomial Expansion

The \((r+1)\text{th}\) term of the binomial expansion of \({(x+y)}^{n}\) is:

\[(\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\]

How To

Given a binomial, write a specific term without fully expanding.

  • Determine the value of \(n\) according to the exponent.
  • Determine \((r+1).\)
  • Determine \(r.\)
  • Replace \(r\) in the formula for the \((r+1)\text{th}\) term of the binomial expansion.
Example 3

Find the tenth term of \({(x+2y)}^{16}\) without fully expanding the binomial.

Use the general term formula \(\binom{n}{r}{x}^{n-r}{y}^{r}\), matching r+1 to the term number you need.

Because we are looking for the tenth term, \(r+1=10,\) we will use \(r=9\) in our calculations.

\[(\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\]

\[(\begin{array}{l}16 \\ 9\end{array}){x}^{16-9}{(2y)}^{9}=5\text{,}857\text{,}280{x}^{7}{y}^{9}\]

Try It #3

Find the sixth term of \({(3x-y)}^{9}\) without fully expanding the binomial.

\(-10,206{x}^{4}{y}^{5}\)

Did you get it?
Media

Access these online resources for additional instruction and practice with binomial expansion.

Key Equations

Table 4
Binomial Theorem\({(x+y)}^{n}=∑k-0n(\begin{array}{l}n \\ k\end{array}){x}^{n-k}{y}^{k}\)
\((r+1)th\) term of a binomial expansion\((\begin{array}{l}n \\ r\end{array}){x}^{n-r}{y}^{r}\)

Key Concepts

Section Exercises

Verbal

1

What is a binomial coefficient, and how it is calculated?

A binomial coefficient is an alternative way of denoting the combination \(C(n,r).\) It is defined as \((\begin{array}{l}n \\ r\end{array})=\,C(n,r)\,=\frac{n!}{r!(n-r)!}.\)

2

What role do binomial coefficients play in a binomial expansion? Are they restricted to any type of number?

3

What is the Binomial Theorem and what is its use?

The Binomial Theorem is defined as \({(x+y)}^{n}=∑k=0n(\begin{array}{l}n \\ k\end{array}){x}^{n-k}{y}^{k}\) and can be used to expand any binomial.

4

When is it an advantage to use the Binomial Theorem? Explain.

Algebraic

For the following exercises, evaluate the binomial coefficient.

5

\((\begin{array}{l}6 \\ 2\end{array})\)

15

6

\((\begin{array}{l}5 \\ 3\end{array})\)

7

\((\begin{array}{l}7 \\ 4\end{array})\)

35

8

\((\begin{array}{l}9 \\ 7\end{array})\)

9

\((\begin{array}{l}10 \\ 9\end{array})\)

10

10

\((\begin{array}{l}25 \\ 11\end{array})\)

11

\((\begin{array}{l}17 \\ 6\end{array})\)

12,376

12

\((\begin{array}{l}200 \\ 199\end{array})\)

For the following exercises, use the Binomial Theorem to expand each binomial.

13

\({(4a-b)}^{3}\)

\(64{a}^{3}-48{a}^{2}b+12a{b}^{2}-{b}^{3}\)

14

\({(5a+2)}^{3}\)

15

\({(3a+2b)}^{3}\)

\(27{a}^{3}+54{a}^{2}b+36a{b}^{2}+8{b}^{3}\)

16

\({(2x+3y)}^{4}\)

17

\({(4x+2y)}^{5}\)

\(1024{x}^{5}+2560{x}^{4}y+2560{x}^{3}{y}^{2}+1280{x}^{2}{y}^{3}+320x{y}^{4}+32{y}^{5}\)

18

\({(3x-2y)}^{4}\)

19

\({(4x-3y)}^{5}\)

\(1024{x}^{5}-3840{x}^{4}y+5760{x}^{3}{y}^{2}-4320{x}^{2}{y}^{3}+1620x{y}^{4}-243{y}^{5}\)

20

\({(\frac{1}{x}+3y)}^{5}\)

21

\({({x}^{-1}+2{y}^{-1})}^{4}\)

\(\frac{1}{{x}^{4}}+\frac{8}{{x}^{3}y}+\frac{24}{{x}^{2}{y}^{2}}+\frac{32}{x{y}^{3}}+\frac{16}{{y}^{4}}\)

22

\({(\sqrt{x}-\sqrt{y})}^{5}\)

For the following exercises, use the Binomial Theorem to write the first three terms of each binomial.

23

\({(a+b)}^{17}\)

\({a}^{17}+17{a}^{16}b+136{a}^{15}{b}^{2}\)

24

\({(x-1)}^{18}\)

25

\({(a-2b)}^{15}\)

\({a}^{15}-30{a}^{14}b+420{a}^{13}{b}^{2}\)

26

\({(x-2y)}^{8}\)

27

\({(3a+b)}^{20}\)

\(3,486,784,401{a}^{20}+23,245,229,340{a}^{19}b+73,609,892,910{a}^{18}{b}^{2}\)

28

\({(2a+4b)}^{7}\)

29

\({({x}^{3}-\sqrt{y})}^{8}\)

\({x}^{24}-8{x}^{21}\sqrt{y}+28{x}^{18}y\)

For the following exercises, find the indicated term of each binomial without fully expanding the binomial.

30

The fourth term of \({(2x-3y)}^{4}\)

31

The fourth term of \({(3x-2y)}^{5}\)

\(-720{x}^{2}{y}^{3}\)

32

The third term of \({(6x-3y)}^{7}\)

33

The eighth term of \({(7+5y)}^{14}\)

\(220,812,466,875,000{y}^{7}\)

34

The seventh term of \({(a+b)}^{11}\)

35

The fifth term of \({(x-y)}^{7}\)

\(35{x}^{3}{y}^{4}\)

36

The tenth term of \({(x-1)}^{12}\)

37

The ninth term of \({(a-3{b}^{2})}^{11}\)

\(1,082,565{a}^{3}{b}^{16}\)

38

The fourth term of \(\,{({x}^{3}-\frac{1}{2})}^{10}\)

39

The eighth term of \(\,{(\frac{y}{2}+\frac{2}{x})}^{9}\)

\(\frac{1152{y}^{2}}{{x}^{7}}\)

Graphical

For the following exercises, use the Binomial Theorem to expand the binomial \(f(x)={(x+3)}^{4}.\) Then find and graph each indicated sum on one set of axes.

40

Find and graph \({f}_{1}(x),\) such that \({f}_{1}(x)\) is the first term of the expansion.

41

Find and graph \({f}_{2}(x),\) such that \({f}_{2}(x)\) is the sum of the first two terms of the expansion.

\({f}_{2}(x)={x}^{4}+12{x}^{3}\)

Graph of the function f_2.
42

Find and graph \({f}_{3}(x),\) such that \({f}_{3}(x)\) is the sum of the first three terms of the expansion.

43

Find and graph \({f}_{4}(x),\) such that \({f}_{4}(x)\) is the sum of the first four terms of the expansion.

\({f}_{4}(x)={x}^{4}+12{x}^{3}+54{x}^{2}+108x\)

Graph of the function f_4.
44

Find and graph \({f}_{5}(x),\) such that \({f}_{5}(x)\) is the sum of the first five terms of the expansion.

Extensions

45

In the expansion of \({(5x+3y)}^{n},\) each term has the form \((\begin{array}{l}n \\ k\end{array}){a}^{n-k}{b}^{k}\) , where \(k\) successively takes on the value \(0,1,2,\,...,\,n.\) If \((\begin{array}{l}n \\ k\end{array})=(\begin{array}{l}7 \\ 2\end{array}),\) what is the corresponding term?

\(590,625{x}^{5}{y}^{2}\)

46

In the expansion of \({(a+b)}^{n},\) the coefficient of \({a}^{n-k}{b}^{k}\) is the same as the coefficient of which other term?

47

Consider the expansion of \(\,{(x+b)}^{40}.\) What is the exponent of \(b\) in the \(k\text{th}\) term?

\(k-1\)

48

Find \(\,(\begin{array}{l}n \\ k-1\end{array})+(\begin{array}{l}n \\ k\end{array})\) and write the answer as a binomial coefficient in the form \(\,(\begin{array}{l}n \\ k\end{array}).\) Prove it. Hint: Use the fact that, for any integer \(p,\) such that \(p\ge 1,\,p!=p(p-1)!\text{.}\)

49

Which expression cannot be expanded using the Binomial Theorem? Explain.

  • \(({x}^{2}-2x+1)\)
  • \({(\sqrt{a}+4\sqrt{a}-5)}^{8}\)
  • \({({x}^{3}+2{y}^{2}-z)}^{5}\)
  • \({(3{x}^{2}-\sqrt{2{y}^{3}})}^{12}\)

The expression \({({x}^{3}+2{y}^{2}-z)}^{5}\) cannot be expanded using the Binomial Theorem because it cannot be rewritten as a binomial.

Glossary

binomial coefficient
the number of ways to chooser objects from n objects where order does not matter; equivalent to \(C(n,r),\) denoted \((\begin{array}{l}n \\ r\end{array})\)
binomial expansion
the result of expanding \({(x+y)}^{n}\) by multiplying
Binomial Theorem
a formula that can be used to expand any binomial