14.8: The Binomial Theorem

Definitions and Theorems

The Binomial Theorem gives a closed formula for the expansion of \((a+b)^n\) for any nonnegative integer \(n\). Before stating it, you need the counting numbers that serve as its coefficients.

Definition: Factorial

For a positive integer \(n\), the factorial of \(n\), written \(n!\), is the product\[n! = n \cdot (n-1) \cdot (n-2) \cdots 3 \cdot 2 \cdot 1.\nonumber\]By definition, \(0! = 1\).

Definition: Binomial Coefficient

For integers \(n\) and \(j\) satisfying \(0 \leq j \leq n\), 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 binomial coefficient \(\binom{n}{j}\) counts the number of \(j\)-element subsets that can be chosen from a set of \(n\) elements. For this reason it is called a combination, and it is read "\(n\) choose \(j\)."

Notations: Combinations

Three notations for the same quantity appear across textbooks and calculators:\[\binom{n}{j}, \quad {}_{n}C_{j}, \quad \text{and} \quad C(n,j).\nonumber\]This text uses \(\binom{n}{j}\) throughout. On most graphing calculators the operation is labeled nCr.

Theorem: Properties of Binomial Coefficients

Let \(n\) and \(j\) be integers with \(0 \leq j \leq n\).

1. \(\binom{n}{0} = \binom{n}{n} = 1\).
2. \(\binom{n}{1} = \binom{n}{n-1} = n\).
3. \(\binom{n}{j} = \binom{n}{n-j}\) (the coefficients are symmetric).
4. \(\binom{n}{j}\) is a positive integer.

Theorem: Pascal's Rule

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

Proof

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

Definition: Pascal's Triangle

Pascal's Triangle is the triangular array whose entry in row \(n\), position \(j\), is \(\binom{n}{j}\), with rows and positions both numbered beginning at \(0\). Each row begins and ends with \(1\), and every interior entry is the sum of the two entries immediately above it.

The first six rows of Pascal's Triangle are\[\begin{array}{ccccccccccc} & & & & & 1 & & & & & \\ & & & & 1 & & 1 & & & & \\ & & & 1 & & 2 & & 1 & & & \\ & & 1 & & 3 & & 3 & & 1 & & \\ & 1 & & 4 & & 6 & & 4 & & 1 & \\ 1 & & 5 & & 10 & & 10 & & 5 & & 1 \end{array}\nonumber\]where the apex is row \(0\). The rule for generating each interior entry is exactly Pascal's Rule.

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 \(n\).

The expansion of \((a+b)^n\) has \(n+1\) terms. In each term the exponents on \(a\) and \(b\) sum to \(n\); the exponent on \(a\) decreases from \(n\) to \(0\) while the exponent on \(b\) increases from \(0\) to \(n\). The coefficients are the entries of row \(n\) of Pascal's Triangle.

Theorem: General Term of a Binomial Expansion

Let \(n\) be a nonnegative integer and let \(k\) be an integer with \(0 \leq k \leq n\). The term of \((a+b)^n\) containing \(b^{k}\) is\[\binom{n}{k} a^{n-k} b^{k}.\nonumber\]This is the \((k+1)\)st term of the expansion, counted from the left.

Caution: Indexing Begins at Zero

The summation index \(k\) starts at \(0\), not at \(1\). Consequently the term produced by \(k=3\) is the fourth term of the expansion, not the third. When a problem asks for "the fifth term," you must use \(k=4\).

Caution: Subtraction and Coefficients

The Binomial Theorem applies to a sum. To expand a difference, rewrite it as a sum before identifying \(a\) and \(b\):\[(2x-5)^3 = \left(2x + (-5)\right)^3, \quad \text{so } a = 2x \text{ and } b = -5.\nonumber\]The entire quantity \(2x\), including its coefficient, is raised to the power \(n-k\), and the entire quantity \(-5\), including its sign, is raised to the power \(k\). Omitting either the coefficient or the sign is the most common source of error in these expansions.

Examples

Example \(\PageIndex{1}\): Evaluating Binomial Coefficients

Evaluate each binomial coefficient.

1. \(\binom{9}{4}\)
2. \(\binom{15}{0}\)
3. \(\binom{12}{10}\)

Solutions

1. Apply the definition with \(n=9\) and \(k=4\). Expand \(9!\) only as far as \(5!\) so that the factor of \(5!\) in the denominator cancels.\[\begin{aligned} \binom{9}{4} &= \dfrac{9!}{4! \, 5!} \\ &= \dfrac{9 \cdot 8 \cdot 7 \cdot 6 \cdot 5!}{4! \, 5!} \\ &= \dfrac{9 \cdot 8 \cdot 7 \cdot 6}{4 \cdot 3 \cdot 2 \cdot 1} \\ &= \dfrac{3024}{24} \\ &= 126 \end{aligned}\nonumber\]
2. By the Properties of Binomial Coefficients, \(\binom{n}{0} = 1\) for every \(n\). Hence \(\binom{15}{0} = 1\). Directly, \(\dfrac{15!}{0! \, 15!} = \dfrac{15!}{1 \cdot 15!} = 1\), using \(0! = 1\).
3. Rather than expanding \(12!\), use symmetry: \(\binom{12}{10} = \binom{12}{12-10} = \binom{12}{2}\). Then\[\binom{12}{2} = \dfrac{12!}{2! \, 10!} = \dfrac{12 \cdot 11}{2 \cdot 1} = \dfrac{132}{2} = 66.\nonumber\]

Example \(\PageIndex{2}\): Expanding with Pascal's Triangle

Extend Pascal's Triangle through row \(6\) and use it to expand \((x+y)^6\).

Solution

Row \(5\) is \(1,\ 5,\ 10,\ 10,\ 5,\ 1\). Row \(6\) begins and ends with \(1\), and each interior entry is the sum of the two entries above it:\[1, \quad 1+5=6, \quad 5+10=15, \quad 10+10=20, \quad 10+5=15, \quad 5+1=6, \quad 1.\nonumber\]So row \(6\) is\[1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1.\nonumber\]These seven numbers are the coefficients of the seven terms of \((x+y)^6\). Attach to them the powers of \(x\) decreasing from \(6\) to \(0\) and the powers of \(y\) increasing from \(0\) to \(6\):\[(x+y)^6 = x^6 + 6x^5y + 15x^4y^2 + 20x^3y^3 + 15x^2y^4 + 6xy^5 + y^6.\nonumber\]As a check, the exponents in every term sum to \(6\).

Example \(\PageIndex{3}\): Expanding a Binomial with a Constant Term

Expand \((x+3)^4\).

Solution

Apply the Binomial Theorem with \(a = x\), \(b = 3\), and \(n = 4\). Row \(4\) of Pascal's Triangle supplies the coefficients \(1,\ 4,\ 6,\ 4,\ 1\).\[(x+3)^4 = \binom{4}{0}x^4 + \binom{4}{1}x^3(3) + \binom{4}{2}x^2(3)^2 + \binom{4}{3}x(3)^3 + \binom{4}{4}(3)^4\nonumber\]Evaluate each power of \(3\) and multiply.\[\begin{aligned} (x+3)^4 &= 1 \cdot x^4 + 4 \cdot x^3 \cdot 3 + 6 \cdot x^2 \cdot 9 + 4 \cdot x \cdot 27 + 1 \cdot 81 \\ &= x^4 + 12x^3 + 54x^2 + 108x + 81 \end{aligned}\nonumber\]

Example \(\PageIndex{4}\): Expanding a Difference with a Leading Coefficient

Expand \((2x-5)^3\).

Solution

Rewrite the difference as a sum so that the theorem applies directly:\[(2x-5)^3 = \left(2x + (-5)\right)^3,\nonumber\]so \(a = 2x\), \(b = -5\), and \(n = 3\). Row \(3\) of Pascal's Triangle supplies the coefficients \(1,\ 3,\ 3,\ 1\).\[(2x-5)^3 = \binom{3}{0}(2x)^3 + \binom{3}{1}(2x)^2(-5) + \binom{3}{2}(2x)(-5)^2 + \binom{3}{3}(-5)^3\nonumber\]Raise each grouped quantity to its power, keeping the coefficient \(2\) and the sign of \(-5\) inside the parentheses.\[\begin{aligned} (2x-5)^3 &= 1 \cdot 8x^3 + 3 \cdot 4x^2 \cdot (-5) + 3 \cdot 2x \cdot 25 + 1 \cdot (-125) \\ &= 8x^3 - 60x^2 + 150x - 125 \end{aligned}\nonumber\]The signs alternate, which is expected whenever \(b\) is negative.

Example \(\PageIndex{5}\): Finding a Single Coefficient

Find the coefficient of \(x^{7}\) in the expansion of \((3x-2)^{10}\) without carrying out the full expansion.

Solution

Write \((3x-2)^{10} = \left(3x + (-2)\right)^{10}\), so \(a = 3x\), \(b = -2\), and \(n = 10\). By the General Term of a Binomial Expansion, the term containing \(b^{k}\) is\[\binom{10}{k}(3x)^{10-k}(-2)^{k}.\nonumber\]The only power of \(x\) in this term comes from \((3x)^{10-k}\), which contributes \(x^{10-k}\). Set that exponent equal to \(7\):\[10-k = 7 \quad \Longrightarrow \quad k = 3.\nonumber\]Substitute \(k=3\) and evaluate.\[\begin{aligned} \binom{10}{3}(3x)^{7}(-2)^{3} &= \dfrac{10!}{3! \, 7!} \cdot 3^{7}x^{7} \cdot (-8) \\ &= 120 \cdot 2187 x^{7} \cdot (-8) \\ &= -2{,}099{,}520\,x^{7} \end{aligned}\nonumber\]The coefficient of \(x^{7}\) is \(-2{,}099{,}520\). Note that this is the fourth term of the expansion, since \(k\) begins at \(0\).

Example \(\PageIndex{6}\): Finding a Constant Term

Find the constant term in the expansion of \(\left(x^{2} - \dfrac{3}{x}\right)^{9}\), where \(x \neq 0\).

Solution

Take \(a = x^{2}\), \(b = -\dfrac{3}{x}\), and \(n = 9\). The general term is\[\binom{9}{k}\left(x^{2}\right)^{9-k}\left(-\dfrac{3}{x}\right)^{k}.\nonumber\]Separate the constants from the powers of \(x\), using \(\dfrac{1}{x} = x^{-1}\):\[\begin{aligned} \binom{9}{k}\left(x^{2}\right)^{9-k}\left(-\dfrac{3}{x}\right)^{k} &= \binom{9}{k}x^{18-2k}\cdot(-3)^{k}x^{-k} \\ &= \binom{9}{k}(-3)^{k}x^{18-3k} \end{aligned}\nonumber\]A constant term is a term whose power of \(x\) is \(0\), so set the exponent equal to \(0\) and solve for \(k\):\[18-3k = 0 \quad \Longrightarrow \quad k = 6.\nonumber\]Since \(6\) is an integer between \(0\) and \(9\), such a term exists. Substitute \(k=6\).\[\begin{aligned} \binom{9}{6}(-3)^{6}x^{0} &= \dfrac{9!}{6! \, 3!}\cdot 729 \\ &= 84 \cdot 729 \\ &= 61{,}236 \end{aligned}\nonumber\]The constant term is \(61{,}236\). Had the equation \(18-3k=0\) produced a non-integer or a value outside \([0,9]\), the expansion would have contained no constant term.

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Sources

Several parts of this text use modifications from the following sources:

• Wikipedia article: "Binomial Theorem"
• Wikipedia article: "Pascal's Triangle"
• Wikipedia article: "Pascal's Rule"

All of these sources are released under the Creative Commons Attribution-Share-Alike License 4.0.