7.6: The Binomial Theorem
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 28661
A binomial is a polynomial with exactly two terms. The binomial theorem gives a formula for expanding \((x+y)^n\) for any positive integer \(n\).
How do we expand a product of polynomials? We pick one term from the first polynomial, multiply by a term chosen from the second polynomial, and then multiply by a term selected from the third polynomial, and so forth. In the special case of \((x+y)^n\), we are selecting either \(x\) or \(y\) from each of the \(n\) binomials \(x+y\) to form a product. Some of these products will be identical, hence, we need to collect their coefficients. The expansion of \((x+y)^3\) is demonstrated below.
We find \[\begin{aligned} (x+y)^3 &= (x+y)(x+y)(x+y) \\[4pt] &= xxx+xxy+xyx+xyy+yxx+yxy+yyx+yyy \\[4pt] &= x^3+x^2y+x^2y+xy^2+x^2y+xy^2+xy^2+y^3 \\[4pt] &= x^3+3x^2y+3xy^2+y^3. \end{aligned}\] What happens when we expand \((x+y)^n\)?
If we select \(y\) from \(k\) copies of the \((x+y)\)s, and \(x\) from the other \(nk\) copies, their product will be \(x^{nk} y^k\). Therefore, in the expansion of \((x+y)^n\), a typical term will be of the form \(x^{nk} y^k\), where \(0\leq k\leq n\). The question is, what is its coefficient in the expansion, after we collect like terms? This coefficient is the number of times the product \(x^{nk} y^k\) appears when we multiply out \((x+y)^n\) in the way described above. It depends on which \(k\) copies of the \((x+y)\)s we will choose \(y\) from. There are \(\binom{n}{k}\) choices, hence, the product \(x^{nk}y^k\) appears \(\binom{n}{k}\) times. Thus, the coefficient is \(\binom{n}{k}\). For this reason, we also call \(\binom{n}{k}\) the binomial coefficients.
Theorem \(\PageIndex{1}\) (Binomial Theorem)
For any positive integer \(n\), \[\begin{aligned} (x+y)^n &= \sum_{k=0}^n \binom{n}{k} x^{nk}y^k. \qquad \end{aligned}\]
Because of the symmetry in the formula, we can interchange \(x\) and \(y\). In addition, we also have \(\binom{n}{k} = \binom{n}{nk}\). Consequently, the binomial theorem can be written in three other forms:
\[\begin{aligned} (x+y)^n &= \sum_{k=0}^n \binom{n}{nk} x^{nk} y^k, \\ (x+y)^n &= \sum_{k=0}^n \binom{n}{ k} x^k y^{nk}, \\ (x+y)^n &= \sum_{k=0}^n \binom{n}{nk} x^k y^{nk}. \end{aligned}\]
You need not worry which one to use. They are all the same! This is how to remember these four different forms. In each term, the powers of \(x\) and \(y\) always add up to \(n\). If the power of one of the two variables is \(k\), where \(0\leq k\leq n\), then the power of the other must be \(nk\), and we need to multiply the coefficient \(\binom{n}{k}\), which is the same as \(\binom{n}{nk}\), to their product.
When expanding \((x+y)^n\), it may be helpful if you first lay out all the terms \(x^n\), \(x^{n1}y\), \(x^{n2}y^2\), and so forth. Then you fill in with the binomial coefficients. For instance, to expand \((x+y)^3\), we first list all the terms that we expect fo find:
\[(x+y)^3 = \underline{\text{ }}\, x^3 + \underline{\text{ }}\, x^2y + \underline{\text{ }}\, xy^2 + \underline{\text{ }}\, y^3. \nonumber\]
Next we fill in the binomial coefficients:
\[(x+y)^3 = \binom{3}{0} x^3 + \binom{3}{1} x^2 y + \binom{3}{2} xy^2 + \binom{3}{3} y^3.\nonumber\]
Finally, evaluate the binomial coefficients and simplify the result.
\[(x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3.\nonumber\]
In a similar way, we also find \((xy)^3 = x^3  3x^2y + 3xy^2  y^3\). Note the similarity between the two expansions.
Example \(\PageIndex{1}\label{eg:binom01}\)
Compute \((x+y)^4\).
 Solution

Following the steps we outlined above, we find
\[\begin{aligned} (x+y)^4 &= \binom{4}{0}x^4 + \binom{4}{1}x^3y + \binom{4}{2}x^2y^2 + \binom{4}{3}xy^3 + \binom{4}{4}y^4 \\ &= x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4. \end{aligned}\]
Since \(\binom{n}{0}=\binom{n}{n}=1\), the expansion always starts with \(x^n\) and ends with \(y^n\).
Example \(\PageIndex{2}\label{eg:binom02}\)
Compute \((xy)^4\).
 Solution

We find
\[\begin{aligned} (xy)^4 &= [x+(y)]^4 \\ &= \binom{4}{0} x^4 + \binom{4}{1}x^3(y) + \binom{4}{2}x^2(y)^2 + \binom{4}{3}x(y)^3 + \binom{4}{4} (y)^4 \\ &= x^4  4x^3y + 6x^2y^2  4xy^3 + y^4. \end{aligned}\]
Take note of the alternating signs in the expansion. This suggests that we could expand \((AB)^n\) the exact same way we would with \((A+B)^n\), except that the signs alternate.
We can carry out the expansion by following these steps. First, list all the terms we expect to find
\[(x+y)^4 = \underline{\text{ }}\, x^4 \phantom{} \underline{\text{ }}\, x^3y \phantom{+} \underline{\text{ }}\, x^2y^2 \phantom{} \underline{\text{ }}\, xy^3 \phantom{+} \underline{\text{ }}\, y^4.\]
Next, fill in the signs:
\[(x+y)^4 = \underline{\text{ }}\, x^4  \underline{\text{ }}\, x^3y + \underline{\text{ }}\, x^2y^2  \underline{\text{ }}\, xy^3 + \underline{\text{ }}\, y^4,\]
and then the binomial coefficients:
\[(x+y)^4 = \binom{4}{0} x^4  \binom{4}{1} x^3y + \binom{4}{2} x^2y^2  \binom{4}{3} xy^3 + \binom{4}{4} y^4.\]
Finally, compute the binomial coefficients to finish the expansion.
Example \(\PageIndex{3}\label{eg:binom03}\)
Expand \((2x3y)^5\).
 Solution

The expansion yields \[(2x)^5\binom{5}{1}(2x)^4(3y)+\binom{5}{2}(2x)^3(3y)^2 \binom{5}{3}(2x)^2(3y)^3+\binom{5}{4}(2x)(3y)^4(3y)^5.\] Therefore, \((2x3y)^5 = 32x^5240x^4y+720x^3y^21080x^2y^3+810xy^4243y^5\).
handson Exercise \(\PageIndex{1}\label{he:binom01}\)
Use the binomial theorem to expand \((3x5y)^4\).
Example \(\PageIndex{4}\label{eg:binom04}\)
Find the coefficient of \(x^3\) in the expansion of \((1+x)^{102}\).
 Solution

Since \[(1+x)^{102} = \sum_{k=0}^{102} \binom{102}{k} x^k,\] the term containing \(x^3\) is \(\binom{102}{3} x^3\). Therefore, the coefficient is \(\binom{102}{3}\). Depending on which form of the binomial theorem you use, you may end up with the term \(\binom{102}{99} x^3\). Numerically, this gives us the same coefficient, because \(\binom{102}{99}=\binom{102}{10299}=\binom{102}{3}\).
\(\binom{102}{3}=171700.\)
Example \(\PageIndex{5}\label{eg:binom05}\)
What is the coefficient of \(t^4\) in the expansion of \((2+3t)^9\)?
 Solution

Since \[(2+3t)^9 = \sum_{k=0}^9 \binom{9}{k} 2^{9k} (3t)^k,\] we need \(k=4\). The coefficient is \(\binom{9}{4} 2^5\cdot3^4=126 \cdot 32 \cdot 81=326,592\).
Example \(\PageIndex{6}\label{eg:binom06}\)
What is the coefficient of \(t^5\) in the expansion of \((32t)^7\)?
 Solution

Since \((32t)^7 = \sum_{k=0}^7 \binom{7}{k} 3^{7k} (2t)^k\), we need \(k=5\), and the coefficient is \(\binom{7}{5}3^2\cdot(2)^5 = \binom{7}{5} 3^2\cdot2^5= 21 \cdot 9 \cdot 32 = 6048\).
handson Exercise \(\PageIndex{2}\label{he:binom02}\)
What is the coefficient of \(t^5\) in \((1+3t)^8\)?
handson Exercise \(\PageIndex{3}\label{he:binom03}\)
What is the coefficient of \(t^4\) in the expansion of \((25t)^9\)?
Example \(\PageIndex{7}\label{eg:binom07}\)
What is the coefficient of \(t^6\) in the expansion of \((4+5t^2)^8\)?
 Solution

The general term in the expansion is \(\binom{8}{k} 4^{8k} (5t^2)^k = \binom{8}{k} 4^{8k} \cdot 5^k t^{2k}\). Hence, we need \(k=3\), and the coefficient is \(\binom{8}{3}4^5\cdot5^3=56 \cdot 1024\cdot 125 = 7,168,000\).
handson Exercise \(\PageIndex{4}\label{he:binom04}\)
What is the coefficient of \(t^9\) in the expansion of \((32t^3)^8\)?
The constant term in an expansion does not contain any variable. It can be interpreted as the term containing \(x^0\).
Example \(\PageIndex{8}\label{eg:binom08}\)
What is the term with \(y^3\) in \(\left(3x+5y\right)^8?\)
 Solution

The general term in the expansion is \[\binom{8}{k} (3x)^{8k} \left(5y\right)^k.\]
Since \(k=3\), the term is \(\binom{8}{3} (3x)^5 \left(5y\right)^3.\)
This is \(56 \cdot (243)x^5 \cdot \left(125 \right) y^3.\)Therefore, the term is \(1701000x^5y^3\).
Pascal's Triangle
(See an introduction to Pascal's Triangle in section 7.4)
To compute the binomial coefficients quickly, one may use the Pascal triangle, in which the \(n\)th row (\(n \geq 0\)) consists of the binomial coefficients \(\binom{n}{k}\), where \(0 \leq k \leq n\):
\[\begin{array}{*{13}{c}} & & & & & & 1 \\ & & & & & 1 & & 1 \\ & & & & 1 & & 2 & & 1 \\ & & & 1 & & 3 & & 3 & & 1 \\ & & 1 & & 4 & & 6 & & 4 & & 1 \\ & 1 & & 5 & &10 & &10 & & 5 & & 1 \\ 1 & & 6 & &15 & &20 & &15 & & 6 & & 1 \end{array}\]
Constructing the Pascal triangle is easy. We generate the rows one at a time. The extreme ends are always 1. Each of the interior entries is the sum of the two entries right above it in the preceding row. For instance, the next row (for \(n=7\)) should be
\[\begin{array}{*{15}{c}} 1 & & 7 & &21 & &35 & &35 & &21 & & 7 & & 1 \end{array}\]
Such computations produce the right binomial coefficients, because of the next result.
Theorem \(\PageIndex{2}\) (Pascal's Identitity)
For all integers \(n\) and \(k\) satisfying \(1\leq k\leq n\), \[\binom{n}{k} = \binom{n1}{k} + \binom{n1}{k1}. \nonumber\]
 (Analytic Proof)

It follows from the definition of binomial coefficients that
\[\begin{aligned} \binom{n1}{k1} + \binom{n1}{k} &= \frac{(n1)!}{(k1)!\,(nk)!} + \frac{(n1)!}{k!\,(nk1)!} \\ &= \frac{(n1)!}{(k1)!\,(nk1)!} \left( \frac{1}{nk} + \frac{1}{k} \right) \\ &= \frac{(n1)!}{(k1)!\,(nk1)!} \cdot \frac{n}{k(nk)} \\ &= \frac{n!}{k!\,(nk)!}.\end{aligned}\]
This completes the proof.
 (Combinatorial Proof)

Let \(A\) be an \(n\)element set. Then \(\binom{n}{k}\) counts the number of \(k\)element subsets of \(A\). These subsets can be classified according to whether they contain a fixed element, say \(x\). If a subset contains \(x\), then the other \(k1\) elements must be selected from the remaining \(n1\) elements of \(A\). Otherwise, if the subset does not contain \(x\), then all its \(k\) elements must be selected from the other \(n1\) elements of \(A\). The numbers of these two kinds of subsets are given by \(\binom{n1}{k1}\) and \(\binom{n1}{k}\), respectively. The theorem now follows immediately by applying the addition principle.
handson Exercise \(\PageIndex{7}\label{he:binom07}\)
Determine the 8th and the 9th rows in the Pascal’s triangle.
Example \(\PageIndex{10}\label{eg:binom10}\)
Use the Pascal’s triangle to expand
 \((CD)^5\)
 \((2A+5B)^3\)
 \((3C4B)^4\)
 Solution

Draw the values of \(\binom{n}{k}\) from the Pascal triangle directly. The answers are:
 \((CD)^5 = C^55C^4D+10C^3D^210C^2D^3+5CD^4D^5\).
 \((2a+5B)^3 = 8A^3+60A^2B+150AB^2+125B^3\).
 \((3C4B)^4 = 81C^4432C^3B+864C^2B^2768CB^3+256B^4\).
Summary and Review
 The binomial theorem can be expressed in four different but equivalent forms.
 The expansion of \((x+y)^n\) starts with \(x^n\), then we decrease the exponent in \(x\) by one, meanwhile increase the exponent of \(y\) by one, and repeat this until we have \(y^n\).
 The next few terms are therefore \(x^{n1}y\), \(x^{n2}y^2\), etc., which end with \(y^n\).
 In general, the sum of exponents in \(x\) and \(y\) is always \(n\). Hence, the general term is \(x^k y^{nk}\), whose coefficient is \(\binom{n}{k}\).
 The expansion of \((x+y)^n\) and \((xy)^n\) look almost identical, except that the signs in \((xy)^n\) alternate.
Exercises
Exercise \(\PageIndex{1}\label{ex:binom01}\)
Expand \((x2y)^4\)
 Answer

\(x^48x^3y+24x^2y^232xy^3+16y^4\)
Exercise \(\PageIndex{2}\label{ex:binom02}\)
Find the coefficient of \(x^{11}y^3\) in \((x+y)^{14}.\)
Exercise \(\PageIndex{3}\label{ex:binom03}\)
Find the coefficient of \(x^4y^7\) in \((2xy)^{11}.\)
 Answer

The term is \(\binom{11}{4}2^4 \cdot x^4(1)^7y^7\), so the coefficient is \(330\cdot 16\cdot 1=5280.\)
Exercise \(\PageIndex{4}\label{ex:binom04}\)
Expand \((3ab)^5\)
Exercise \(\PageIndex{5}\label{ex:binom05}\)
Find the coefficient of \(y^5\) in \((2x+3y)^7.\)
 Answer

The term is \(\binom{7}{5}2^2 \cdot x^2\cdot 3^5y^5\), so the coefficient is \(21\cdot 4\cdot 243=20,412.\)
Exercise \(\PageIndex{6}\label{ex:binom06}\)
Find the coefficient of \(y^3\) in \((5x2y)^5.\)
Exercise \(\PageIndex{7}\label{ex:binom07}\)
What is the term in \((2x+7y)^6\) with \(y^2?\)
 Answer

The term is \(\binom{6}{2}2^4 \cdot x^4\cdot 7^2y^2\), so the term is \(15\cdot 16\cdot 49x^4y^2=11760x^4y^2\)
Exercise \(\PageIndex{8}\label{ex:binom08}\)
What is the term in \((x3y)^8\) with \(x^5?\)