14.6: Binomial Theorem

Learning Objectives Simplify expressions involving factorials. Understand and use binomial coefficients and the Binomial Theorem to expand the powers of binomials quickly.

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Recall that a polynomial with two terms is called a binomial. We have already learned how 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.

In this section, we aim to prove the celebrated Binomial Theorem. Simply stated, the Binomial Theorem is a formula for the expansion of quantities \((a+b)^n\) for natural numbers \(n\). In Elementary and Intermediate Algebra, you should have seen specific instances of the formula, namely\[\begin{array}{rcl} (a+b)^1 & = & a + b \\[6pt] (a+b)^2 & = & a^2 + 2ab + b^2 \\[6pt] (a+b)^3 & = & a^3 + 3a^2 b + 3ab^2 + b^3 \\[6pt] \end{array}\nonumber\]If we wanted the expansion for \((a+b)^4\), we would write \((a+b)^4 = (a+b)(a+b)^3\) and use the formula that we have for \((a+b)^3\) to get\[(a+b)^4 = (a+b) \left( a^3 + 3a^2 b + 3ab^2 + b^3 \right) = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4.\nonumber\]Generalizing this a bit, we see that if we have a formula for \((a+b)^{k}\), we can obtain a formula for \((a+b)^{k+1}\) by rewriting the latter as\[(a+b)^{k+1} = (a+b)(a+b)^{k}.\nonumber\]This means Mathematical Induction plays a significant role in the proof of the Binomial Theorem.

Factorials

Before we can state the theorem, we need to revisit the sequence of factorials that were introduced earlier in this chapter.

Definition: Factorial (Recursive Definition)

The factorial of a natural number \( n \), denoted \( n! \) and reads as "\( n \) factorial," is defined by the recursive formula\[ \begin{array}{rclclcc}
a_0 & = & 0! & = & 1 & & \\[6pt]
a_n & = & n! & = & n \cdot a_{n- 1} & \quad & \text{for } n \geq 1 \\[6pt]
\end{array}  \nonumber \]

Recall this means \(0! = 1\) and \(n! = n(n-1)!\) for \(n \geq 1\). Using the recursive definition, we get: \(1! = 1 \cdot 0! = 1 \cdot 1 = 1\), \(2! = 2 \cdot 1! = 2 \cdot 1 = 2\), \(3! = 3 \cdot 2! = 3 \cdot 2 \cdot 1 = 6\), and \(4! = 4 \cdot 3! = 4 \cdot 3 \cdot 2 \cdot 1 = 24\). Informally, \(n! = n\cdot(n -1)\cdot(n -2) \cdots 2 \cdot 1\) with \(0! = 1\) as our "base case." Our first example refamiliarizes us with some of the basic computations involving factorials.

Example \( \PageIndex{1} \): Computing Factorials

1. Simplify the following expressions.
   1. \(\frac{3! \, 2!}{0!}\)
   2. \(\frac{7!}{5!}\)
   3. \(\frac{1000!}{998! \, 2!}\)
   4. \(\frac{(k+2)!}{(k-1)!}\), \(k \geq 1\)
2. Prove \(n! \gt 3^n\) for all \(n \geq 7\).

Solutions

1. We keep in mind the mantra, "When in doubt, write it out!" as we simplify the following.

   1. We have been programmed to react with alarm to the presence of a \(0\) in the denominator, but in this case, \(0! = 1\), so the fraction is defined after all. As for the numerator, \(3! = 3 \cdot 2 \cdot 1 = 6\) and \(2! = 2 \cdot 1 = 2\), so we have \(\frac{3! \, 2!}{0!} = \frac{(6)(2)}{1} = 12\).
   2. We have \(7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040\) while \(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120\). Dividing, we get \(\frac{7!}{5!} = \frac{5040}{120} = 42\). While this is correct, we note that we could have saved ourselves some time had we proceeded as follows\[\dfrac{7!}{5!} = \dfrac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = \dfrac{7 \cdot 6 \cdot \cancel{5} \cdot \cancel{4} \cdot \cancel{3} \cdot \cancel{2} \cdot \cancel{1}}{\cancel{5} \cdot \cancel{4} \cdot \cancel{3} \cdot \cancel{2} \cdot \cancel{1}} = 7 \cdot 6 = 42.\nonumber\]In fact, should we want to exploit the recursive nature of the factorial fully, we can write\[\dfrac{7!}{5!} = \dfrac{7 \cdot 6 \cdot 5!}{5!} = \dfrac{7 \cdot 6 \cdot \cancel{5!}}{\cancel{5!}} = 42.\nonumber\]
   3. Keeping in mind the lesson we learned from the previous problem, we have\[\dfrac{1000!}{998! \, 2!} = \dfrac{1000 \cdot 999 \cdot 998!}{998! \cdot 2!} = \dfrac{1000 \cdot 999 \cdot \cancel{998!}}{\cancel{998!} \cdot 2!} = \dfrac{999000}{2} = 499500\nonumber\]
   4. This problem continues the theme we have seen in the previous two problems. We first note that since \(k+2\) is larger than \(k-1\), \((k+2)!\) contains all of the factors of \((k-1)!\) and as a result we can get the \((k-1)!\) to cancel from the denominator. To see this, we begin by writing out \((k+2)!\) starting with \((k+2)\) and multiplying it by the numbers which precede it until we reach \((k-1)\): \((k+2)! = (k+2)(k+1)(k)(k-1)!\). As a result, we have\[\dfrac{(k+2)!}{(k-1)!} = \dfrac{(k+2)(k+1)(k)(k-1)!}{(k-1)!} = \dfrac{(k+2)(k+1)(k) \cancel{(k-1)!}}{\cancel{(k-1)!}} = k(k+1)(k+2)\nonumber\]The stipulation \(k \geq 1\) is there to ensure that all of the factorials involved are defined.

2. This will require a proof by Mathematical Induction. In this case, the claim should only hold for \( n \geq 7 \). Let \( P(n) \) be the statement \( n! \gt 3^n \).

   For \( n = 7 \), we have \( 7! = 5040 \gt 2187 = 3^7 \). Thus, \( P(7) \) is true.

   Induction Hypothesis: Assume \( P(k) \) is true. That is, assume \( k! \gt 3^k \).

   We want to show that \( P(k + 1) \) is true. That is, we want to show \( (k + 1)! \gt 3^{k + 1} = 3 \cdot 3^k \).\[ \begin{array}{rclr} (k + 1)! & = & (k + 1)k! & \left( \text{Definition of a factorial} \right) \\[6pt] & \gt & (k + 1)3^k & \left( \text{Induction Hypothesis} \right) \\[6pt] & \gt & (3)3^k & \left( k+1 \gt 3 \text{ when }k \gt 2 \text{. Luckily, we know that }k \geq 7\text{, so }k+1 \gt 3 \right) \\[6pt] & = & 3^{k + 1} & \\[6pt] \end{array} \nonumber \]Hence, we have shown \( (k + 1)! \gt 3^{k + 1} \). Thus, \(P(k+1)\) is true. By the Principle of Mathematical Induction, we have \(n! > 3^{n}\) for all \(n \geq 7\).

Of all the mathematical animals we discussed in the text, factorials grow most quickly. In Example \( \PageIndex{1b} \), we proved that \(n!\) overtakes \(3^{n}\) at \(n=7\). "Overtakes" may be too polite a word since \(n!\) thoroughly trounces \(3^n\) for \(n \geq 7\), as any reasonable set of data will show. It can be shown that for any real number \(x > 0\), not only does \(n!\) eventually overtake \(x^n\), but the ratio \(\frac{x^n}{n!} \to 0\) as \(n \to \infty\). This fact is far more important than you might realize right now, but wait until Calculus and you might begin to understand the significance of this statement.

Binomial Coefficients

Applications of factorials in the wild often involve counting arrangements. For example, suppose you have fifty songs on your phone and wish to arrange these songs in a playlist in which the order of the songs matters. There are \(50!\) different possible playlists in that case. That equates to \( 30,414,093,201,713,378,043,612,608,166,064,768,844,377,641,568,960,512,000,000,000,000 \) possible playlists. If you wish to select only ten of the songs to create a playlist, then there are \(\frac{50!}{40!} = 37,276,043,023,296,000\) such playlists. If, on the other hand, you want to select ten song files out of the fifty to play in any order, there are \(\frac{50!}{40! 10!} = 10,272,278,170\) ways to achieve this. The authors encourage you to take courses such as Finite Mathematics, Discrete Mathematics, and Statistics if you are interested in concepts like this. We introduce these concepts here because this is how the factorials make their way into the Binomial Theorem, as the following definition indicates.

Definition: Binomial Coefficient

Given two whole numbers \(n\) and \(j\) with \(n \geq j\), the binomial coefficient \(\displaystyle \binom{n}{j}\) (read, "\(n\) choose \(j\)") is the whole number given by\[\binom{n}{j} = \dfrac{n!}{j! (n-j)!}.\nonumber\]

The name "binomial coefficient" will be justified shortly. For now, we can physically interpret \(\binom{n}{j}\) as the number of ways to select \(j\) items from \(n\) items where the order of the items selected is unimportant. For example, suppose you win two free tickets to a special screening of the latest Hollywood blockbuster and have five friends, each of whom would love to accompany you to the movies. There are \(\binom{5}{2}\) ways to choose who goes with you. Applying the definition of the binomial coefficient, we get\[\binom{5}{2} = \dfrac{5!}{2! (5-2)!} = \dfrac{5!}{2! 3!} = \dfrac{5 \cdot 4}{2} = 10.\nonumber\]So there are \(10\) different ways to distribute those two tickets among five friends. (Some will see it as \(10\) ways to decide which three friends must stay home.) The reader is encouraged to verify this by listing all the possibilities.

Example \( \PageIndex{ 2 } \): Finding the Number of Combinations Using the Formula

A fast food restaurant offers five side dish options. Your meal comes with two side dishes.

1. How many ways can you select your side dishes?
2. How many ways can you select 3 side dishes?

Solutions

1. We want to choose 2 side dishes from 5 options.\[\binom{5}{2} = \dfrac{5!}{2!(5−2)!} = \dfrac{5 \cdot 4 \cdot \cancel{3!}}{2! \cdot \cancel{3!}} = \dfrac{20}{2} = 10. \nonumber \]There are ten ways to choose 2 side dishes from 5 options.
2. We want to choose 3 side dishes from 5 options.\[\binom{5}{3} = \dfrac{5!}{3!(5−3)!} = \dfrac{5 \cdot 4 \cdot \cancel{3!}}{\cancel{3!} \cdot 2!} = \dfrac{20}{2} = 10. \nonumber \]Therefore, there are also ten ways to choose 3 side dishes from 5 options.

The equivalence of the answers to Example \( \PageIndex{ 2a } \) and \( \PageIndex{ 2b } \) is not a coincidence. Asking how many ways to choose 2 items from 5 should be the same as asking how to exclude 3 items from 5.

Checkpoint \( \PageIndex{ 2 } \)

An ice cream shop offers 10 flavors of ice cream. How many ways are there to choose 3 flavors for a banana split?

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 \( \PageIndex{ 3 } \): Finding Binomial Coefficients

Find each binomial coefficient.

1. \( \binom{5}{3} \)
2. \( \binom{9}{2} \)
3. \( \binom{9}{7} \)

Solutions

Use the formula to calculate each binomial coefficient:\[ \binom{n}{r} = \dfrac{n!}{r!(n−r)!}. \nonumber \]

1. \( \binom{5}{3} = \frac{5!}{3!(5−3)!} = \frac{5 \cdot 4 \cdot \cancel{3!}}{\cancel{3!} \cdot 2!} = \frac{20}{2} = 10 \)
2. \( \binom{9}{2} = \frac{9!}{2!(9−2)!} = \frac{9 \cdot 8 \cdot \cancel{7!}}{2! \cdot \cancel{7!}} = \frac{72}{2} = 36 \)
3. \( \binom{9}{7} = \frac{9!}{7!(9−7)!} = \frac{9 \cdot 8 \cdot \cancel{7!}}{\cancel{7!} \cdot 2!} = \frac{72}{2} = 36 \)

Checkpoint \( \PageIndex{ 3 } \)

Find each binomial coefficient.

1. \( \binom{7}{3} \)
2. \( \binom{11}{4} \)

The Binomial Theorem

We now state and prove a theorem, named after the celebrated French mathematician Blaise Pascal, which is crucial to the proof of the Binomial Theorem.

Theorem: Pascal's Rule

For natural numbers \(n\) and \(j\) with \(n \geq j\),\[\binom{n}{j-1} + \binom{n}{j} = \binom{n+1}{j}.\nonumber\nonumber\]

Proof

\[\begin{array}{rcl} \displaystyle{\binom{n}{j-1} + \binom{n}{j}} & = & \dfrac{n!}{(j-1)! (n-(j-1))!} + \dfrac{n!}{j! (n-j)!} \\[6pt] & = & \dfrac{n!}{(j-1)! (n-j+1)!} + \dfrac{n!}{j! (n-j)!} \\[6pt] & = & \dfrac{n!}{(j-1)! (n-j+1)(n-j)!} + \dfrac{n!}{j(j-1)! (n-j)!} \\[6pt] & = & \dfrac{n! \, j}{j(j-1)! (n-j+1)(n-j)!} + \dfrac{n! (n-j+1)}{j(j-1)! (n-j+1)(n-j)!} \\[6pt] & = & \dfrac{n! \, j}{j! (n-j+1)!} + \dfrac{n! (n-j+1)}{j! (n-j+1)!} \\[6pt] & = & \dfrac{n! \, j + n! (n-j+1)}{j! (n-j+1)!} \\[6pt] & = & \dfrac{n!\left( j + (n-j+1)\right)}{j! (n-j+1)!} \\[6pt] & = & \dfrac{(n+1) n!}{j! (n+1-j))!} \\[6pt] & = & \dfrac{(n+1)!} {j! ((n+1)-j))!} \\[6pt] & = & \displaystyle{\binom{n+1}{j}} \, \checkmark \\[6pt] \end{array}\nonumber\]

We can now state and prove the Binomial Theorem. This theorem states that binomial coefficients are just that - coefficients in the binomial expansion.

Theorem: Binomial Theorem

For nonzero real numbers \(a\) and \(b\),\[(a+b)^{n} =\displaystyle{\sum_{j=0}^{n} \binom{n}{j} a^{n-j} b^{j}}\nonumber\]for all natural numbers \(n\).

Proof

To prove the Binomial Theorem, we let \(P(n)\) be the expansion formula given in the statement of the theorem, and we note that \(P(1)\) is true since\[\begin{array}{rcl} (a+b)^{1} & \stackrel{?}{=} & \displaystyle{\sum_{j=0}^{1} \binom{1}{j} a^{1-j} b^{j}} \\[6pt] a+b & \stackrel{?}{=} & \displaystyle{\binom{1}{0}a^{1-0}b^{0} + \binom{1}{1}a^{1-1}b^{1}} \\[6pt] a+b & = & a + b \, \checkmark \\[6pt] \end{array}\nonumber\]Now we assume that \(P(k)\) is true. That is, we assume that we can expand \((a+b)^k\) using the formula given in the Binomial Theorem and attempt to show that \(P(k+1)\) is true.\[\begin{array}{rcl} (a+b)^{k+1} & = & (a+b)(a+b)^{k} \\[6pt] & = & (a+b) \displaystyle{\sum_{j=0}^{k} \binom{k}{j} a^{k-j} b^{j}} \\[6pt] & = & a \displaystyle{\sum_{j=0}^{k} \binom{k}{j} a^{k-j} b^{j}} + b \displaystyle{\sum_{j=0}^{k} \binom{k}{j} a^{k-j} b^{j}} \\[6pt] & = & \displaystyle{\sum_{j=0}^{k} \binom{k}{j} a^{k+1-j} b^{j}} + \displaystyle{\sum_{j=0}^{k} \binom{k}{j} a^{k-j} b^{j+1}} \\[6pt] \end{array}\nonumber\]We aim to combine as many terms as possible within the two summations. As the counter \(j\) in the first summation runs from \(0\) through \(k\), we get terms involving \(a^{k+1}\), \(a^{k}b\), \(a^{k-1}b^2\), \ldots , \(ab^{k}\). In the second summation, we get terms involving \(a^{k}b\), \(a^{k-1}b^{2}\), \ldots , \(ab^{k}\), \(b^{k+1}\). In other words, they have common terms apart from the first term in the first summation and the last term in the second summation. Our next move is to "kick out" the terms that we cannot combine and rewrite the summations so that we can combine them. To that end, we note\[\displaystyle{\sum_{j=0}^{k} \binom{k}{j} a^{k+1-j} b^{j} = a^{k+1}+ \sum_{j=1}^{k} \binom{k}{j} a^{k+1-j} b^{j}}\nonumber\]and\[\displaystyle{\sum_{j=0}^{k} \binom{k}{j} a^{k-j} b^{j+1} = \sum_{j=0}^{k-1} \binom{k}{j} a^{k-j} b^{j+1} + b^{k+1}}\nonumber\]so that\[(a+b)^{k+1} = \displaystyle{a^{k+1} + \sum_{j=1}^{k} \binom{k}{j} a^{k+1-j} b^{j} + \sum_{j=0}^{k-1} \binom{k}{j} a^{k-j} b^{j+1} + b^{k+1}}.\nonumber\]We now wish to write\[\displaystyle{\sum_{j=1}^{k} \binom{k}{j} a^{k+1-j} b^{j} + \sum_{j=0}^{k-1} \binom{k}{j} a^{k-j} b^{j+1}}\nonumber\]as a single summation. The wrinkle is that the first summation starts with \(j=1\), while the second starts with \(j=0\). Even though the sums produce terms with the same powers of \(a\) and \(b\), they do so for different values of \(j\). To resolve this, we need to shift the index on the second summation so that the index \(j\) starts at \(j=1\) instead of \(j=0\).\[\begin{array}{rcl} \displaystyle{ \sum_{j=0}^{k-1} \binom{k}{j} a^{k-j} b^{j+1}} & = & \displaystyle{\sum_{j=0+1}^{k-1+1} \binom{k}{j-1} a^{k-(j-1)} b^{(j-1)+1}} \\[6pt] & = & \displaystyle{\sum_{j=1}^{k} \binom{k}{j-1} a^{k+1-j} b^{j}} \\[6pt] \end{array}\nonumber\]We can now combine our two sums and simplify.\[\begin{array}{rcl} \displaystyle{\sum_{j=1}^{k} \binom{k}{j} a^{k+1-j} b^{j} + \sum_{j=0}^{k-1} \binom{k}{j} a^{k-j} b^{j+1}} & = & \displaystyle{\sum_{j=1}^{k} \binom{k}{j} a^{k+1-j} b^{j} + \sum_{j=1}^{k} \binom{k}{j-1} a^{k+1-j} b^{j}} \\[6pt] & = & \displaystyle{\sum_{j=1}^{k} \left[ \binom{k}{j} + \binom{k}{j-1} \right] a^{k+1-j} b^{j} } \\[6pt] & = & \displaystyle{\sum_{j=1}^{k} \binom{k+1}{j} a^{k+1-j} b^{j} } \\[6pt] \end{array}\nonumber\]Using this and the fact that \(\binom{k+1}{0} = 1\) and \(\binom{k+1}{k+1} = 1\), we get\[\begin{array}{rcl} (a+b)^{k+1} & = & a^{k+1} + \displaystyle{\sum_{j=1}^{k} \binom{k+1}{j} a^{k+1-j} b^{j} } + b^{k+1} \\[6pt] & = & \displaystyle{ \binom{k+1}{0} a^{k+1} b^{0} + \sum_{j=1}^{k} \binom{k+1}{j} a^{k+1-j} b^{j} + \binom{k+1}{k+1} a^{0} b^{k+1}} \\[6pt] & = & \displaystyle{ \sum_{j=0}^{k+1} \binom{k+1}{j} a^{(k+1)-j} b^{j}} \\[6pt] \end{array}\nonumber\]which shows that \(P(k+1)\) is true. Hence, by induction, we have established that the Binomial Theorem holds for all natural numbers \(n\).

To get a feel of what this theorem is saying and how it is easier to remember than it may first appear, let's consider the specific case of \(n=4\). According to the theorem, we have\[\begin{array}{rcl} (a+b)^{4} & = & \displaystyle{\sum_{j=0}^{4} \binom{4}{j} a^{4-j} b^{j}} \\[6pt] & = & \displaystyle{\binom{4}{0}a^{4-0}b^{0} + \binom{4}{1}a^{4-1}b^{1} + \binom{4}{2}a^{4-2}b^{2} + \binom{4}{3}a^{4-3}b^{3} + \binom{4}{4}a^{4-4}b^{4}} \\[6pt] & = & \displaystyle{\binom{4}{0}a^{4} + \binom{4}{1}a^{3}b + \binom{4}{2}a^{2}b^{2} + \binom{4}{3}ab^{3} + \binom{4}{4}b^{4}} \\[6pt] \end{array}\nonumber\]We forgo the simplification of the coefficients to note the pattern in the expansion. First, note that in each term, the total of the exponents is \(4\), which matches the exponent of the binomial \((a+b)^{4}\). The exponent on \(a\) begins at \(4\) and decreases by one as we move from one term to the next, while the exponent on \(b\) starts at \(0\) and increases by one each time. Also, note that the binomial coefficients themselves have a pattern. The upper number, \(4\), matches the exponent on the binomial \((a+b)^4\), whereas the lower number changes from term to term and matches the exponent of \(b\) in that term. This is no coincidence and corresponds to the kind of counting we discussed earlier. If we think of obtaining \((a+b)^4\) by multiplying \((a+b)^2(a+b)^2\), our answer is the sum of all possible products with exactly four factors - some \(a\), some \(b\). If we wish to count, for instance, the number of ways we obtain \(1\) factor of \(b\) out of a total of \(4\) possible factors, thereby forcing the remaining \(3\) factors to be \(a\), the answer is \(\binom{4}{1}\). Hence, the term \(\binom{4}{1}a^{3}b\) is in the expansion. The other terms which appear cover the remaining cases.

Example \( \PageIndex{4} \): Evaluating Binomials Using the Binomial Theorem

Use the Binomial Theorem to find the following.

1. \((x-2)^4\)
2. \(2.1^{3}\)
3. The term containing \(x^3\) in the expansion \((2x+y)^{5}\)

Solutions

1. Since \((x-2)^4 = (x+(-2))^4\), we identify \(a = x\), \(b = -2\) and \(n=4\) and obtain\[\begin{array}{rcl} (x-2)^4\! & = & \displaystyle{\sum_{j=0}^{4} \binom{4}{j} x^{4-j} (-2)^{j}} \\[6pt] & = & \displaystyle{\binom{4}{0} x^{4-0} (-2)^{0} \!+\! \binom{4}{1} x^{4-1} (-2)^{1} \!+\! \binom{4}{2} x^{4-2} (-2)^{2} \!+\! \binom{4}{3} x^{4-3} (-2)^{3} \!+\! \binom{4}{4} x^{4-4} (-2)^{4}} \\[6pt] & = & x^4 -8x^3 + 24x^2 - 32x + 16 \\[6pt] \end{array}\nonumber\]
2. At first, this problem seems misplaced, but we can write \(2.1^{3} = (2 + 0.1)^3\). Identifying \(a =2\), \(b = 0.1 = \frac{1}{10}\) and \(n=3\), we get\[\begin{array}{rcl} \left(2+\dfrac{1}{10} \right)^3 & = & \displaystyle{\sum_{j=0}^{3} \binom{3}{j} 2^{3-j} \left(\dfrac{1}{10}\right)^{j}} \\[6pt] & = & \displaystyle{\binom{3}{0} 2^{3-0} \left(\dfrac{1}{10}\right)^{0} + \binom{3}{1} 2^{3-1} \left(\dfrac{1}{10}\right)^{1} + \binom{3}{2} 2^{3-2} \left(\dfrac{1}{10}\right)^{2} + \binom{3}{3} 2^{3-3} \left(\dfrac{1}{10}\right)^{3}} \\[6pt] & = & 8 + \dfrac{12}{10} + \dfrac{6}{100} + \dfrac{1}{1000} \\[6pt] & = & 8 + 1.2 + 0.06 + 0.001 \\[6pt] & = & 9.261 \\[6pt] \end{array}\nonumber\]
3. Identifying \(a = 2x\), \(b = y\) and \(n=5\), the Binomial Theorem gives\[(2x+y)^{5} = \displaystyle{\sum_{j=0}^{5} \binom{5}{j} (2x)^{5-j} y^{j}}\nonumber\]Since we are concerned with only the term containing \(x^3\), there is no need to expand the entire sum. The exponents on each term must add to \(5\) and if the exponent on \(x\) is \(3\), the exponent on \(y\) must be \(2\). Plucking out the term \(j=2\), we get\[\displaystyle{\binom{5}{2} (2x)^{5-2} y^{2}} = 10 (2x)^3y^2 = 80x^3y^2\nonumber\]

Checkpoint \( \PageIndex{ 4 } \)

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

Pascal's Triangle

We close this section with Pascal’s Triangle. Pascal's Triangle is obtained by arranging the binomial coefficients in the triangular fashion below.\[\begin{array}{ccccccccc} & & & & \displaystyle{\binom{0}{0}} & & & & \\[6pt] & & & \displaystyle{\binom{1}{0}} & & \displaystyle{\binom{1}{1}} & & &\\[6pt] & & & & \searrow \, \swarrow & & & & \\[6pt] & & \displaystyle{\binom{2}{0}} & & \displaystyle{\binom{2}{1}} & & \displaystyle{\binom{2}{2}} & &\\[6pt] & & & \searrow \, \swarrow & & \searrow \, \swarrow & & & \\[6pt] & \displaystyle{\binom{3}{0}} & & \displaystyle{\binom{3}{1}} & & \displaystyle{\binom{3}{2}} & & \displaystyle{\binom{3}{3}} & \\[6pt] & & \searrow \, \swarrow & & \searrow \, \swarrow & & \searrow \, \swarrow & & \\[6pt] \displaystyle{\binom{4}{0}} & & \displaystyle{\binom{4}{1}} & & \displaystyle{\binom{4}{2}} & & \displaystyle{\binom{4}{3}} & & \displaystyle{\binom{4}{4}} \\[6pt] & & & & \vdots & & & & \\[6pt] \end{array}\nonumber\]Since \(\binom{n}{0} = 1\) and \(\binom{n}{n} = 1\) for all whole numbers \(n\), we get that each row of Pascal's Triangle begins and ends with \(1\). We use the additive relationship expressed in our first theorem in this section to generate the numbers in the middle of the rows (from the third row onwards). For instance, \(\binom{1}{0} + \binom{1}{1} = \binom{2}{1}\), \(\binom{2}{0} + \binom{2}{1} = \binom{3}{1}\), and so forth. The arrows in the array above indicate this relationship. With these two facts, we can quickly generate Pascal's Triangle. We start with the first two rows, \(1\) and \(1 \quad 1\). From that point on, each successive row begins and ends with \(1\), and the middle numbers are generated using the first theorem in this section. Below we attempt to demonstrate this building process to generate the first five rows of Pascal’s Triangle.\[\begin{array}{ccc} \begin{array}{ccccccccc} & & & & 1 & & & & \\[6pt] & & & 1 & & 1 & & &\\[6pt] & & & & \searrow \, \swarrow & & & & \\[6pt] & & \fbox{1} & & \underline{1+1} & & \fbox{1} & &\\[6pt] \end{array} & \xrightarrow{\hspace{.5in}} & \begin{array}{ccccccccc} & & & & 1 & & & & \\[6pt] & & & 1 & & 1 & & &\\[6pt] & & 1 & & 2 & &1 & &\\[6pt] \end{array} \\[6pt] && \\[6pt] \begin{array}{ccccccccc} & & & & 1 & & & & \\[6pt] & & & 1 & & 1 & & &\\[6pt] & & 1 & & 2 & &1 & &\\[6pt] & & & \searrow \, \swarrow & & \searrow \, \swarrow & & & \\[6pt] & \fbox{1} & & \underline{1+2} & & \underline{2+1} & & \fbox{1} & \\[6pt] \end{array} & \xrightarrow{\hspace{.5in}} & \begin{array}{ccccccccc} & & & & 1 & & & & \\[6pt] & & & 1 & & 1 & & &\\[6pt] & & 1 & & 2 & &1 & &\\[6pt] & 1 & & 3 & & 3 & & 1 & \\[6pt] \end{array} \\[6pt] && \\[6pt] \begin{array}{ccccccccc} & & & & 1 & & & & \\[6pt] & & & 1 & & 1 & & &\\[6pt] & & 1 & & 2 & &1 & &\\[6pt] & 1 & & 3 & & 3 & & 1 & \\[6pt] & & \searrow \, \swarrow & & \searrow \, \swarrow & & \searrow \, \swarrow & & \\[6pt] \fbox{1} & & \underline{1+3} & & \underline{3+3} & & \underline{3+1} & & \fbox{1} \\[6pt] \end{array} & \xrightarrow{\hspace{.5in}} & \begin{array}{ccccccccc} & & & & 1 & & & & \\[6pt] & & & 1 & & 1 & & &\\[6pt] & & 1 & & 2 & &1 & &\\[6pt] & 1 & & 3 & & 3 & & 1 & \\[6pt] 1 & & 4 & & 6 & & 4 & & 1 \\[6pt] \end{array} \\[6pt] \end{array}\nonumber\]To see how we can use Pascal’s Triangle to expedite the Binomial Theorem, suppose we wish to expand \((3x-y)^{4}\). The coefficients we need are \(\binom{4}{j}\) for \(j = 0, 1, 2, 3, 4\) and are the numbers that form the fifth row of Pascal's Triangle. Since we know that the exponent of \(3x\) in the first term is \(4\) and then decreases by one as we go from left to right while the exponent of \(-y\) starts at \(0\) in the first term and then increases by one as we move from left to right, we quickly obtain\[\begin{array}{rcl} (3x-y)^{4} & = & (1)(3x)^{4} + (4)(3x)^3(-y) + (6)(3x)^2(-y)^2 + 4(3x)(-y)^3 + 1(-y)^4 \\[6pt] & = & 81x^4 - 108x^3y + 54x^2y^2 -12xy^3 + y^4 \\[6pt] \end{array}\nonumber\]We would like to stress that Pascal's Triangle is a rapid method to expand an entire binomial. If only a term (or two or three) is required, then the Binomial Theorem is the way to go.

Checkpoint \( \PageIndex{ 5 } \)

Write in expanded form.

1. \((x−y)^5\)
2. \((2x+5y)^3\)