How To Find Nth Term In Binomial Expansion
13.6: Binomial Theorem
- Page ID
- 3348
- Apply the Binomial Theorem.
A polynomial with ii terms is chosen a binomial. We accept already learned to multiply binomials and to heighten binomials to powers, but raising a binomial to a high power can be ho-hum and time-consuming. In this section, nosotros will hash out a shortcut that will allow usa to notice \((x+y)^n\) without multiplying the binomial by itself \(due north\) times.
Identifying Binomial Coefficients
In the shortcut to finding \({(x+y)}^n\), nosotros will need to use combinations to detect the coefficients that will announced in the expansion of the binomial. In this example, we utilize the annotation \(\dbinom{north}{r}\) instead of \(C(north,r)\), but it can be calculated in the same way. So
\[\dbinom{northward}{r}=C(n,r)=\dfrac{due north!}{r!(n−r)!}\]
The combination \(\dbinom{n}{r}\) is called a binomial coefficient. An example of a binomial coefficient is:
\(\dbinom{five}{2}=C(5,2)=10\)
If \(n\) and \(r\) are integers greater than or equal to \(0\) with \(due north≥r\), then the binomial coefficient is
\[\dbinom{due north}{r}=C(north,r)=\dfrac{n!}{r!(n−r)!} \characterization{binomial1}\]
Yep. Simply as the number of combinations must e'er be a whole number, a binomial coefficient will always be a whole number.
Find each binomial coefficient.
- \(\dbinom{v}{3}\)
- \(\dbinom{9}{2}\)
- \(\dbinom{9}{seven}\)
Solution
Use the Equation \ref{binomial1} to summate each binomial coefficient. You can also utilise the \(nC_r\) office on your computer.
- \(\dbinom{v}{3}=\dfrac{v!}{3!(5−3)!}=\dfrac{5⋅4⋅3!}{3!2!}=10\)
- \(\dbinom{nine}{2}=\dfrac{9!}{2!(nine−2)!}=\dfrac{ix⋅8⋅7!}{two!seven!}=36\)
- \(\dbinom{9}{7}=\dfrac{9!}{7!(9−seven)!}=\dfrac{9⋅eight⋅7!}{seven!2!}=36\)
Assay
Detect that we obtained the same issue for parts (b) and (c). If you look closely at the solution for these ii parts, you lot volition see that you cease up with the same two factorials in the denominator, just the order is reversed, simply every bit with combinations.
\[\dbinom{northward}{r}=\dbinom{n}{due north−r} \nonumber\]
Find each binomial coefficient.
- \(\dbinom{vii}{3}\)
- \(\dbinom{eleven}{4}\)
- Answer a
-
\(35\)
- Answer b
-
\(33\)
Using the Binomial Theorem
When we aggrandize \({(x+y)}^due north\) past multiplying, the issue is chosen a binomial expansion, and it includes binomial coefficients. If we wanted to aggrandize \({(x+y)}^{52}\), nosotros might multiply \((x+y)\) by itself fifty-ii times. This could take hours! If we examine some simple binomial expansions, nosotros can notice patterns that will pb us to a shortcut for finding more than complicated binomial expansions.
\[\begin{marshal*} {(x+y)}^two &= x^2+2xy+y^2 \\[4pt] {(10+y)}^3 &= x^3+3x^2y+3xy^2+y^3 \\[4pt] {(x+y)}^4 &= x^iv+4x^3y+6x^2y^2+4xy^3+y^4 \stop{align*}\]
Starting time, permit'south examine the exponents. With each successive term, the exponent for \(x\) decreases and the exponent for \(y\) increases. The sum of the ii exponents is \(n\) for each term.
Next, allow'southward examine the coefficients. Notice that the coefficients increase and and so subtract in a symmetrical pattern. The coefficients follow a blueprint:
\(\dbinom{north}{0}\), \(\dbinom{n}{1}\), \(\dbinom{n}{2}\),..., \(\dbinom{n}{n}.\)
These patterns atomic number 82 usa to the Binomial Theorem, which can exist used to aggrandize any binomial.
\[\begin{marshal*} {(x+y)}^northward&=\sum_{k=0}^{n}\dbinom{n}{k}x^{n−one thousand}y^m \\[4pt] &=x^n+\dbinom{due north}{1}10^{n−ane}y+\dbinom{north}{2}10^{due north−2}y^2+...+\dbinom{due north}{n−1}xy^{n−ane}+y^north \end{marshal*}\]
Some other way to see the coefficients is to examine the expansion of a binomial in general class, \(x+y\), to successive powers \(ane\), \(2\), \(3\), and \(four\).
\[\begin{align*} {(x+y)}^1 &= ten+y \\ {(x+y)}^ii &= x^2+2xy+y^ii \\ {(x+y)}^iii &= x^3+3x^2y+3xy^2+y^iii \\ {(10+y)}^4 &= x^4+4x^3y+6x^2y^two+4xy^3+y^4 \cease{marshal*}\]
Can you lot approximate the next expansion for the binomial \({(x+y)}^5\)?
Figure \(\PageIndex{one}\)
Encounter Figure \(\PageIndex{i}\), which illustrates the following:
- There are \(due north+1\) terms in the expansion of \({(x+y)}^due north\).
- 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 \(y\) brainstorm with \(0\) and increase to \(due north\).
- The coefficients are symmetric.
To decide the expansion on \({(x+y)}^v\), we come across \(north=five\), thus, there will be \(5+1=half dozen\) terms. Each term has a combined degree of \(5\). In descending order for powers of \(10\), the design is as follows:
- Introduce \(ten^5\), and so 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.
\(10^five, x^4y, ten^3y^2, x^2y^3, xy^4, y^5\)
The next expansion would be
\({(x+y)}^five=x^5+5x^4y+10x^3y^2+10x^2y^three+5xy^four+y^5\)
Simply where practice those coefficients come from? The binomial coefficients are symmetric. Nosotros tin see these coefficients in an array known equally Pascal's Triangle, shown in Figure \(\PageIndex{2}\).
Figure \(\PageIndex{2}\)
To generate Pascal'south Triangle, nosotros start past writing a \(i\). In the row below, row 2, we write two \(1's\). In the 3rd row, flank the ends of the rows with \(1'southward\), and add \(one+1\) to notice the middle number, \(2\). In the \(n^{thursday}\) row, flank the ends of the row with \(1's\). Each element in the triangle is the sum of the two elements immediately above information technology.
To come across the connection betwixt Pascal'due south Triangle and binomial coefficients, permit united states revisit the expansion of the binomials in general form.
The Binomial Theorem is a formula that can be used to expand any binomial.
\[ {(ten+y)}^n = \sum_{k=0}^{northward}\dbinom{n}{1000}ten^{n−k}y^k = x^n+\dbinom{n}{ane}x^{n−1}y+\dbinom{northward}{2}x^{n−2}y^2+...+\dbinom{n}{n−1}xy^{n−1}+y^northward \]
- Determine the value of \(northward\) according to the exponent.
- Evaluate the \(k=0\) through \(k=n\) using the Binomial Theorem formula.
- Simplify.
Write in expanded form.
- \({(10+y)}^five\)
- \({(3x−y)}^iv\)
Solution
a. Substitute \(northward=five\) into the formula. Evaluate the \(g=0\) through \(k=5\) terms. Simplify.
\[\brainstorm{align*} {(x+y)}^v &= \dbinom{5}{0}x^5y^0+\dbinom{5}{ane}x^4y^1+\dbinom{5}{2}x^3y^two+\dbinom{5}{3}10^2y^three+\dbinom{5}{iv}x^1y^4+\dbinom{5}{5}10^0y^v \\ {(10+y)}^5 &= ten^5+5x^4y+10x^3y^2+10x^2y^3+5xy^iv+y^5 \end{marshal*}\]
b. Substitute \(n=iv\) into the formula. Evaluate the \(k=0\) through \(thousand=4\) terms. Notice that \(3x\) is in the identify that was occupied by \(ten\) and that \(–y\) is in the place that was occupied by \(y\). So we substitute them. Simplify.
\[\begin{align*} {(3x−y)}^4 &= \dbinom{4}{0}{(3x)}^4{(−y)}^0+\dbinom{4}{i}{(3x)}^3{(−y)}^one+\dbinom{4}{two}{(3x)}^2{(−y)}^2+\dbinom{4}{3}{(3x)}^1{(−y)}^3+\dbinom{4}{four}{(3x)}^0{(−y)}^iv \\ {(3x−y)}^4 &= 81x^4−108x^3y+54x^2y^ii−12xy^3+y^4 \end{align*}\]
Assay
Observe the alternating signs in part b. This happens considering \((−y)\) 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.
- \({(x−y)}^v\)
- \({(2x+5y)}^3\)
- Reply a
-
\(10^5−5x^4y+10x^3y^ii−10x^2y^3+5xy^4−y^5\)
- Reply b
-
\(8x^3+60x^2y+150xy^ii+125y^iii\)
Using the Binomial Theorem to Observe a Single Term
Expanding a binomial with a high exponent such as \({(ten+2y)}^{16}\) tin can exist a lengthy procedure. Sometimes nosotros are interested merely in a certain term of a binomial expansion. We exercise not need to fully expand a binomial to discover a single specific term.
Note the design of coefficients in the expansion of \({(x+y)}^5\).
\({(10+y)}^v=x^5+\dbinom{five}{1}x^4y+\dbinom{5}{two}x^3y^ii+\dbinom{5}{3}x^2y^three+\dbinom{5}{iv}xy^4+y^5\)
The 2d term is \(\dbinom{5}{one}x^4y\). The third term is \(\dbinom{5}{2}x^3y^two\). We can generalize this upshot.
The \((r+1)\)th term of the binomial expansion of \({(x+y)}^northward\) is:
\[\dbinom{due north}{r}x^{n−r}y^r \label{binomial5}\]
- Determine the value of nn co-ordinate to the exponent.
- Determine \((r+1)\).
- Decide \(r\).
- Supersede \(r\) in the formula for the \((r+1)\)th term of the binomial expansion.
Detect the tenth term of \({(x+2y)}^{16}\) without fully expanding the binomial.
Solution
Considering we are looking for the 10th term, \(r+1=10\), we will use \(r=nine\) in our calculations and Equation \ref{binomial5}.
\(\dbinom{16}{ix}ten^{16−9}{(2y)}^9=5,857,280x^7y^9\)
Find the sixth term of \({(3x−y)}^9\) without fully expanding the binomial.
- Answer
-
\(−x,206x^4y^5\)
Key Equations
| Binomial Theorem | \({(10+y)}^n=\sum_{k=0}^northward\dbinom{n}{m}x^{due north−1000}y^m\) |
| \((r+one)\)thursday term of a binomial expansion | \(\dbinom{due north}{r}x^{n−r}y^r\) |
Primal Concepts
- \(\dbinom{n}{r}\) is called a binomial coefficient and is equal to \(C(n,r)\). Come across Example \(\PageIndex{ane}\).
- The Binomial Theorem allows us to aggrandize binomials without multiplying. Meet Example \(\PageIndex{2}\).
- We can discover a given term of a binomial expansion without fully expanding the binomial. Come across Example \(\PageIndex{3}\).
Source: https://math.libretexts.org/Bookshelves/Algebra/Book%3A_Algebra_and_Trigonometry_(OpenStax)/13%3A_Sequences_Probability_and_Counting_Theory/13.06%3A_Binomial_Theorem
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