13.6: Binomial Theorem
- Page ID
- 3348
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- Apply the Binomial Theorem.
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 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 \(\dbinom{n}{r}\) instead of \(C(n,r)\), but it can be calculated in the same way. So
\[\dbinom{n}{r}=C(n,r)=\dfrac{n!}{r!(n−r)!}\]
The combination \(\dbinom{n}{r}\) is called a binomial coefficient. An example of a binomial coefficient is:
\(\dbinom{5}{2}=C(5,2)=10\)
If \(n\) and \(r\) are integers greater than or equal to \(0\) with \(n≥r\), then the binomial coefficient is
\[\dbinom{n}{r}=C(n,r)=\dfrac{n!}{r!(n−r)!} \label{binomial1}\]
Yes. Just as the number of combinations must always be a whole number, a binomial coefficient will always be a whole number.
Find each binomial coefficient.
- \(\dbinom{5}{3}\)
- \(\dbinom{9}{2}\)
- \(\dbinom{9}{7}\)
Solution
Use the Equation \ref{binomial1} to calculate each binomial coefficient. You can also use the \(nC_r\) function on your calculator.
- \(\dbinom{5}{3}=\dfrac{5!}{3!(5−3)!}=\dfrac{5⋅4⋅3!}{3!2!}=10\)
- \(\dbinom{9}{2}=\dfrac{9!}{2!(9−2)!}=\dfrac{9⋅8⋅7!}{2!7!}=36\)
- \(\dbinom{9}{7}=\dfrac{9!}{7!(9−7)!}=\dfrac{9⋅8⋅7!}{7!2!}=36\)
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.
\[\dbinom{n}{r}=\dbinom{n}{n−r} \nonumber\]
Find each binomial coefficient.
- \(\dbinom{7}{3}\)
- \(\dbinom{11}{4}\)
- Answer a
-
\(35\)
- Answer b
-
\(33\)
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.
\[\begin{align*} {(x+y)}^2 &= x^2+2xy+y^2 \\[4pt] {(x+y)}^3 &= x^3+3x^2y+3xy^2+y^3 \\[4pt] {(x+y)}^4 &= x^4+4x^3y+6x^2y^2+4xy^3+y^4 \end{align*}\]
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:
\(\dbinom{n}{0}\), \(\dbinom{n}{1}\), \(\dbinom{n}{2}\),..., \(\dbinom{n}{n}.\)
These patterns lead us to the Binomial Theorem, which can be used to expand any binomial.
\[\begin{align*} {(x+y)}^n&=\sum_{k=0}^{n}\dbinom{n}{k}x^{n−k}y^k \\[4pt] &=x^n+\dbinom{n}{1}x^{n−1}y+\dbinom{n}{2}x^{n−2}y^2+...+\dbinom{n}{n−1}xy^{n−1}+y^n \end{align*}\]
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 \(4\).
\[\begin{align*} {(x+y)}^1 &= x+y \\ {(x+y)}^2 &= x^2+2xy+y^2 \\ {(x+y)}^3 &= x^3+3x^2y+3xy^2+y^3 \\ {(x+y)}^4 &= x^4+4x^3y+6x^2y^2+4xy^3+y^4 \end{align*}\]
Can you guess the next expansion for the binomial \({(x+y)}^5\)?
Figure \(\PageIndex{1}\)
See Figure \(\PageIndex{1}\), which illustrates the following:
- There are \(n+1\) terms in the expansion of \({(x+y)}^n\).
- The degree (or sum of the exponents) for each term is \(n\).
- The powers on \(x\) begin with \(n\) and decrease to \(0\).
- The powers on \(y\) begin with \(0\) and increase to \(n\).
- The coefficients are symmetric.
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:
- Introduce \(x^5\), and then for each successive term reduce the exponent on \(x\) by \(1\) until \(x^0=1\) is reached.
- Introduce \(y^0=1\), and then increase the exponent on yy by 1 until \(y^5\) is reached.
\(x^5, x^4y, x^3y^2, x^2y^3, xy^4, y^5\)
The next expansion would be
\({(x+y)}^5=x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5\)
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 \(\PageIndex{2}\).
Figure \(\PageIndex{2}\)
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 \(n^{th}\) 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.
The Binomial Theorem is a formula that can be used to expand any binomial.
\[ {(x+y)}^n = \sum_{k=0}^{n}\dbinom{n}{k}x^{n−k}y^k = x^n+\dbinom{n}{1}x^{n−1}y+\dbinom{n}{2}x^{n−2}y^2+...+\dbinom{n}{n−1}xy^{n−1}+y^n \]
- Determine the value of \(n\) according to the exponent.
- Evaluate the \(k=0\) through \(k=n\) using the Binomial Theorem formula.
- Simplify.
Write in expanded form.
- \({(x+y)}^5\)
- \({(3x−y)}^4\)
Solution
a. Substitute \(n=5\) into the formula. Evaluate the \(k=0\) through \(k=5\) terms. Simplify.
\[\begin{align*} {(x+y)}^5 &= \dbinom{5}{0}x^5y^0+\dbinom{5}{1}x^4y^1+\dbinom{5}{2}x^3y^2+\dbinom{5}{3}x^2y^3+\dbinom{5}{4}x^1y^4+\dbinom{5}{5}x^0y^5 \\ {(x+y)}^5 &= x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5 \end{align*}\]
b. 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.
\[\begin{align*} {(3x−y)}^4 &= \dbinom{4}{0}{(3x)}^4{(−y)}^0+\dbinom{4}{1}{(3x)}^3{(−y)}^1+\dbinom{4}{2}{(3x)}^2{(−y)}^2+\dbinom{4}{3}{(3x)}^1{(−y)}^3+\dbinom{4}{4}{(3x)}^0{(−y)}^4 \\ {(3x−y)}^4 &= 81x^4−108x^3y+54x^2y^2−12xy^3+y^4 \end{align*}\]
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.
Write in expanded form.
- \({(x−y)}^5\)
- \({(2x+5y)}^3\)
- Answer a
-
\(x^5−5x^4y+10x^3y^2−10x^2y^3+5xy^4−y^5\)
- Answer b
-
\(8x^3+60x^2y+150xy^2+125y^3\)
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 \({(x+y)}^5\).
\({(x+y)}^5=x^5+\dbinom{5}{1}x^4y+\dbinom{5}{2}x^3y^2+\dbinom{5}{3}x^2y^3+\dbinom{5}{4}xy^4+y^5\)
The second term is \(\dbinom{5}{1}x^4y\). The third term is \(\dbinom{5}{2}x^3y^2\). We can generalize this result.
The \((r+1)\)th term of the binomial expansion of \({(x+y)}^n\) is:
\[\dbinom{n}{r}x^{n−r}y^r \label{binomial5}\]
- Determine the value of nn according to the exponent.
- Determine \((r+1)\).
- Determine \(r\).
- Replace \(r\) in the formula for the \((r+1)\)th term of the binomial expansion.
Find the tenth term of \({(x+2y)}^{16}\) without fully expanding the binomial.
Solution
Because we are looking for the tenth term, \(r+1=10\), we will use \(r=9\) in our calculations and Equation \ref{binomial5}.
\(\dbinom{16}{9}x^{16−9}{(2y)}^9=5,857,280x^7y^9\)
Find the sixth term of \({(3x−y)}^9\) without fully expanding the binomial.
- Answer
-
\(−10,206x^4y^5\)
Access these online resources for additional instruction and practice with binomial expansion.
Key Equations
| Binomial Theorem | \({(x+y)}^n=\sum_{k=0}^n\dbinom{n}{k}x^{n−k}y^k\) |
| \((r+1)\)th term of a binomial expansion | \(\dbinom{n}{r}x^{n−r}y^r\) |
Key Concepts
- \(\dbinom{n}{r}\) is called a binomial coefficient and is equal to \(C(n,r)\). See Example \(\PageIndex{1}\).
- The Binomial Theorem allows us to expand binomials without multiplying. See Example \(\PageIndex{2}\).
- We can find a given term of a binomial expansion without fully expanding the binomial. See Example \(\PageIndex{3}\).
Contributors and Attributions
Jay Abramson (Arizona State University) with contributing authors. Textbook content produced by OpenStax College is licensed under a Creative Commons Attribution License 4.0 license. Download for free at https://openstax.org/details/books/precalculus.