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Modeling, Functions, and Graphs

Katherine Yoshiwara

Section 9.5 The Binomial Expansion

Earlier we studied products of polynomials, and in particular we found expanded forms for powers of binomials such as \((a+b)^2\) and \((a+b)^3\text{.}\) The amount of work involved in expanding such powers increases as the exponent gets larger. In this section we will learn how to raise a binomial to any positive integer power, without having to perform the polynomial products.

Investigation 9.4. Powers of Binomials.

In this investigation we will look for patterns in the expansion of \((a+b)^n\text{.}\) We begin by computing several such powers.
Expand each power and fill in the blanks. Arrange the terms in each expansion in descending powers of \(a\text{.}\)
  1. \((a+b)^0 = \)
  2. \((a+b)^1 = \)
  3. \((a+b)^2 = \)
  4. \((a+b)^3 = \)
    Hint: Start by writing
    \begin{equation*} (a+b)^3 = (a+b)(a+b)^2 \end{equation*}
    and use your answer to #3.
  5. \((a+b)^4 = \)
    Hint: Start by writing
    \begin{equation*} (a+b)^4 = (a+b)(a+b)^4 \end{equation*}
    and use your answer to #4.
  6. \((a+b)^5 = \)
  7. Do you see a relationship between the exponent \(n\) and the number of terms in the expansion of \((a+b)^n\text{?}\) (Notice that for \(n=0\) we have \((a+b)^0=1\text{,}\) which has one term.) Fill in the table below.
    \(n\) Number of terms
    in \((a+b)^n\)
    \(0\) \(\hphantom{0000}\)
    \(1\) \(\hphantom{0000}\)
    \(2\) \(\hphantom{0000}\)
    \(3\) \(\hphantom{0000}\)
    \(4\) \(\hphantom{0000}\)
    \(5\) \(\hphantom{0000}\)
  8. First observation: In general, the expansion of \((a+b)^n\) has terms.
  9. Next we’ll consider the exponents on \(a\) and \(b\) in each term of the expansions. Refer to your expanded powers in parts 1-5, and fill in the next table.
    \(n\) First term of
    \((a+b)^n\)     		Last term of
    \((a+b)^n\)     		Sum of exponents
    on \(a\) and \(b\)
    in each term
    \(0\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(1\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(2\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(3\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(4\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    \(5\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
  10. Second observation: In any term of the expansion of \((a+b)^n\text{,}\) the sum of the exponents on \(a\) and on \(b\) is
  11. In fact, we can be more specific in describing the exponents in the expansions. We will use \(k\) to label the terms in the expansion of \((a+b)^n\text{,}\) starting with \(k=0\text{.}\) For example, for \(n=2\) we label the terms as follows: .
    \begin{equation*} \begin{aligned}[t] (a+b)^2 = \amp ~~a^2~~~~+~~~~2ab~~~~+~~~~b^2\\ \amp k=0~~~~~~~~k=1~~~~~~~~k=2 \\ \end{aligned} \end{equation*}
    We can make a table showing the exponents on \(a\) and \(b\) in each term of \((a+b)^2\text{:}\)
    Case \(n=2\text{:}\)
    \(k\) \(0\) \(1\) \(2\)
    Exponent on \(a\) \(2\) \(1\) \(0\)
    Exponent on \(b\) \(0\) \(1\) \(2\)
    Complete the tables shown below for the cases \(n=3,~ n=4\) and \(n=5\text{.}\)
    Case \(n=3\text{:}\)
    \(k\) \(0\) \(1\) \(2\) \(3\)
    Exponent on \(a\) \(3\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    Exponent on \(b\) \(0\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    Case \(n=4\text{:}\)
    \(k\) \(0\) \(1\) \(2\) \(3\) \(4\)
    Exponent on \(a\) \(4\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    Exponent on \(b\) \(0\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    Case \(n=5\text{:}\)
    \(k\) \(0\) \(1\) \(2\) \(3\) \(4\) \(5\)
    Exponent on \(a\) \(4\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
    Exponent on \(b\) \(0\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\) \(\hphantom{0000}\)
  12. Third observation: The variable factors of the \(k^{\text{th}}\) term in the expansion of \((a+b)^n\) may be expressed as . (Fill in the correct powers in terms of \(k\) and \(n\) for \(a\) and \(b\text{.}\))
In the next investigation we will look for patterns in the coefficients of the terms of the expansions.

Subsection Powers of Other Binomials

We can use what we learned from the Investigation to raise other binomials to powers.

Example 9.69.

Expand each power of a binomial.
  1. \(\displaystyle (x+1)^3\)
  2. \(\displaystyle (2m-n)^4\)
Solution.
  1. We know from the Investigation that
    \begin{gather*} (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3 \end{gather*}
    We replace \(a\) with \(x\) and \(b\) with 1 to get
    \begin{align*} (x+1)^3 \amp = x^3 +3x^2(1) + 3x(1)^2 + 1^3\\ \amp = x^3 + 3x^2 + 3x + 1 \end{align*}
  2. This time we take the expansion
    \begin{gather*} (a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4 \end{gather*}
    from the Investigation and let \(a=2m\) and \(b = -n\text{.}\)
    \begin{align*} (2m-n)^4 \amp = (2m)^4 + 4(2m)^3(-n) + 6(2m)^2(-n)^2 + 4(2m)(-n)^3 + (-n)^4\\ \amp = 16m^4 + 4(8m^3)(-n) + 6(4m^2)(n^2) + 4(2m)(-n^3) + n^4\\ \amp = 16m^4 - 32m^3n + 24m^2n^2 - 8mn^3 + n^4 \end{align*}

Checkpoint 9.70. Practice 1.

Expand \((r+2s)^3=\)
Answer.
\(r^{3}+6r^{2}s+12rs^{2}+8s^{3}\)
Solution.
\(r^3 + 6r^2s + 12rs^2 + 8s^3\)

Subsection The Binomial Coefficient

Now we continue our study of powers of binomials. What should the expansion of \((a+b)^6\) look like? From the Iinvestigation we know that it should have 7 terms, with the exponents on \(a\) decreasing from 6 down to 0, and the exponents on \(b\) increasing from 0 up to 6.
Thus, the expansion should have the form
\begin{equation*} (a+b)^6 =\fillinmath{XXX}a^6 + \fillinmath{XXXX}a^5b+\fillinmath{XXX}a^4b^2+\fillinmath{XXXX}a^3b^3+\fillinmath{XXX}a^2b^4+\fillinmath{XXXX}ab^5+\fillinmath{XXX}b^6 \end{equation*}
But what about the numerical coefficients of the terms? We will use the notation \(_nC_k \) for the coefficient of the \(k^{\text{th}}\) term in the expansion of \((a+b)^n\text{.}\) For example, in the expansion of \((a+b)^6\) above,
  • the coefficient of \(a^6\) is denoted by \(_6C_0\text{,}\)
  • the coefficient of \(a^5b^1\) is \(_6C_1 \text{,}\)
  • the coefficient of \(a^4b^2\) is \(_6C_2 \text{,}\)
and so on. Note that the 6 in front of the \(C\) indicates that \(n=6\text{,}\) and the number following \(C\) corresponds to the exponent on \(b\text{.}\) The symbol \(~_nC_k~\) is called the binomial coefficient.

The Binomial Coefficient.

The binomial coefficient \(~_nC_k~\) is the coefficient of the term containing \(~b^k~\) in the expansion of \((a+b)^n\text{.}\)

Example 9.71.

Use your expansions from the Investigation to evaluate the following binomial coefficients.
  1. \(\displaystyle _4C_3 \)
  2. \(\displaystyle _5C_2 \)
Solution.
  1. \(_4C_3\) is the coefficient of the term containing \(b^3\) in the expansion of \((a+b)^4\text{.}\) Referring to Step 5 in the Investigation, we see that the coefficient of the term \(4ab^3\) is 4. Thus, \(_4C_3 = 4\text{.}\)
  2. \(_5C_2\) is the coefficient of the term containing \(b^2\) in the expansion of \((a+b)^5\text{.}\) Referring to Step 6 in the Investigation, we see that the coefficient of the term \(10a^3b^2\) is 10. Thus, \(_5C_2 = 10\text{.}\)

Checkpoint 9.72. Practice 2.

Evaluate the binomial coefficients.
  1. \(_5C_4 =\)
  2. \(_5C_0 =\)
Answer 1.
\(5\)
Answer 2.
\(1\)
Solution.
  1. 5
  2. 1

Investigation 9.5. Pascal’s Triangle.

To get a clearer picture of the binomial coefficients, consider again the expansions of \((a+b)^n\) you calculated in the last Investigation, but this time look only at the numerical coefficients of each term:
\(n=0\) \(\hphantom{0000}\) \(1\)
\(n=1\) \(1\hphantom{000}1\)
\(n=2\) \(1\hphantom{000}2\hphantom{000}1\)
\(n=3\) \(1\hphantom {000}3\hphantom{000} 3 \hphantom{000}1\)
\(n=4\) \(1\hphantom {000}4\hphantom{000} 6 \hphantom{000}4\hphantom{000}1\)
\(n=5\) \(1\hphantom {000} 5\hphantom{ii0} 10 \hphantom{00}10 \hphantom{0ii}5 \hphantom{000}1\)
This triangular array of numbers is known as Pascal’s triangle. It has many interesting and surprising properties that have been extensively studied. We might first make the following observations.
  1. Each row of Pascal’s triangle begins with the number and ends with the number .
  2. The second number and the next-to-last number in the \(n^{\text{th}}\) row are .
    There is an interesting pattern for the rest of the numbers in each row. Pick any number in the row \(n=4\text{,}\) and look at the two closest numbers in the previous row. (For example, if you picked 6, the two closest numbers in the previous row are 3 and 3.) Do you see a relationship between the numbers? Try the same thing with several numbers in row \(n=5\) to test your theory.
  3. Starting with the row \(n=2\text{,}\) any number in the triangle (except the first and last 1’s in each row) can be found by .
  4. Using your answer to question #3, continue Pascal’s triangle to include the row for \(n=6\text{.}\)
    The numbers in Pascal’s triangle are the binomial coefficients we have been looking for. Specifically, the number in the \(k^{\text{th}}\) position (starting with \(k=0\)) of the \(n^{\text{th}}\) row of the triangle is \(~_nC_k~\text{.}\) Thus, we can use the numbers in Pascal’s triangle to expand \((a+b)^n\text{.}\)
  5. Use Pascal’s triangle to find the binomial coefficient \(~_6C_4~\text{.}\)
  6. Expand: \((a+b)^6\) .
  7. Expand: \((x-2)^6\) . (Hint: Use your answer to #6, replacing \(a\) by \(x\) and \(b\) by \(-2\text{.}\))
  8. Continue Pascal’s triangle to include the row for \(n=7\text{.}\)

Subsection Using Pascal’s Triangle

We can now write the expanded form of any binomial power without having to perform the multiplication.

Example 9.73.

Find the expanded form of \((3r-q)^6\) without performing the multiplication.
Solution.
Start by writing down the expansion of \((a+b)^6\text{.}\)
\begin{gather*} (a+b)^6 = a^6 + \fillinmath{XXX}a^5b+\fillinmath{XXX}a^4b^2+\fillinmath{XXX}a^3b^3+\fillinmath{XXX}+a^2b^4+\fillinmath{XXX}ab^5+b^6 \end{gather*}
Next, fill in the blanks with the binomial coefficients from Pascal’s triangle.
\begin{gather*} (a+b)^6 = a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \end{gather*}
Finally, replace \(a\) by \(3r\) and replace \(b\) by \(-q\text{,}\) then simplify to find
\begin{align*} (3r-q)^6 \amp = (3r)^6 + 6(3r)^5(-q) + 15(3r)^4(-q)^2 + 20(3r)^3(-q)^3\\ \amp\qquad\qquad + 15(3r)^2(-6)^4 + 6(3r)(-q)^5 + (-q)^6\\ \amp = 729r^6 - 1458r^5q + 1215r^4q^2 - 540r^3q^3 + 135r^2q^2 - 18rq^5 + q^6 \end{align*}

Checkpoint 9.74. Practice 3.

Expand \((w+10z)^6\text{.}\)
Answer.
\(w^{6}+60w^{5}z+1500w^{4}z^{2}+20000w^{3}z^{3}+150000w^{2}z^{4}+600000wz^{5}+1\times 10^{6}z^{6}\)
Solution.
\(w^6 + 60w^5z+1500w^4z^2+20,000w^3z^3+150,000w^2z^4+{}\)\(600,000wz^5+1,000,000z^6\)

Subsection Factorial Notation

Finding binomial coefficients from Pascal’s triangle is tedious for large values of \(n\text{.}\) It would be convenient to have a formula for calculating the coefficients directly. For this formula we need a new symbol, \(n!\) (read "\(n\) factorial"), which is defined as follows:

Factorial.

If \(n\) is a positive integer,
\begin{equation*} n! = n \cdot (n-1) \cdot (n-2) \cdot \cdots \cdot 3\cdot 2 \cdot 1 \end{equation*}
For example,
\begin{equation*} 6! = 5 \cdot 4\cdot 3\cdot 2\cdot 1 = 720 ~~~~~\text{and}~~~~~ 4! = 4\cdot 3\cdot 2\cdot 1 = 24 \end{equation*}

Note 9.75.

The factorial symbol applies only to the variable or number it follows; for example, \(3 \cdot 4!\) is not equal to \((3 \cdot 4)!\text{.}\)

Example 9.76.

Write each expression in expanded form and simplify.
  1. \(2n!\) for \(n=4\)
  2. \((2n-1)!\) for \(n=4\)
Solution.
  1. \(\displaystyle 2n! = 2 \cdot 4! = 2(4 \cdot 3 \cdot 2 \cdot 1) = 48\)
  2. \(\displaystyle (2n-1)! = 7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 5040\)

Checkpoint 9.77. Practice 4.

Evalute \(4 \cdot 3!=\)
Answer.
\(24\)
Solution.
24

Example 9.78.

Write \((3n+1)!\) in factored form, showing the first three factors and the last three factors.
Solution.
\((3n+1)! = (3n+1)\cdot (3n) \cdot (3n-1) \cdot \cdots \cdot 3 \cdot 2 \cdot 1\)

Checkpoint 9.79. Practice 5.

Write the first three factors of
  1. \(2n!\)
    • \(\displaystyle 2\cdot n \cdot (n-1) \)
    • \(\displaystyle 2n\cdot (2n-2) \cdot (2n-4) \)
    • \(\displaystyle 2n\cdot (2n-1) \cdot (2n-2) \)
  2. \((2n)!\)
    • \(\displaystyle 2\cdot n \cdot (n-1) \)
    • \(\displaystyle 2n\cdot (2n-2) \cdot (2n-4) \)
    • \(\displaystyle 2n\cdot (2n-1) \cdot (2n-2) \)
Answer 1.
\(\text{Choice 1}\)
Answer 2.
\(\text{Choice 3}\)
Solution.
  1. \(\displaystyle 2 \cdot n \cdot (n-1)\)
  2. \(\displaystyle (2n) \cdot (2n-1) \cdot (2n-2)\)
Note that \(7! = 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\) can be written as
\begin{equation*} 7 \cdot 6! = 7 \cdot (6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1) \end{equation*}
and \(5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1\) can be written as
\begin{equation*} 5 \cdot 4! = 5 \cdot (4 \cdot 3 \cdot 2 \cdot 1) \end{equation*}
In general, we have the following relationship:
\begin{equation*} n! = n\cdot(n-1)! \end{equation*}
This relationship can be helpful in simplifying expressions involving factorials.

Example 9.80.

Write each expression in expanded form and simplify.
  1. \(\displaystyle \dfrac{6!}{3!}\)
  2. \(\displaystyle \dfrac{4!~6!}{8!}\)
Solution.
  1. \(\displaystyle \dfrac{6!}{3!} = \dfrac{6 \cdot 5 \cdot 4 \cdot 3!}{3!} = 6 \cdot 5 \cdot 4 = 120\)
  2. \(\displaystyle \dfrac{4!~6!}{8!} = \dfrac{4\cdot 3 \cdot 2 \cdot 1\cdot 6!}{8\cdot 7 \cdot 6!} = \dfrac{3}{7}\)

Caution 9.81.

In the previous Example, note that \(\dfrac{6!}{3!}\) is not equal to \(2!\text{.}\)

Checkpoint 9.82. Practice 6.

Write \(\dfrac{5!}{3!~2!}\) in expanded form and simplify.
Simplified form:
Answer.
\(10\)
Solution.
\(10\)
We have defined \(n!\) when \(n\) is a positive integer. What if \(n=0\text{?}\) To be consistent with the equation \(n! = n\cdot(n-1)!\) we must have
\begin{equation*} 1! = 1 \cdot (1-1)!~~~~~~~~\text{or} ~~~~~~~~1! = 1 \cdot (0)! \end{equation*}
This leads us to the following definition.

Zero Factorial.

\begin{equation*} 0! = 1 \end{equation*}
Note that both \(1!\) and \(0!\) are equal to \(1\text{.}\)

Subsection The Binomial Coefficient in Factorial Notation

Earlier we introduced the notation \(~_nC_k~\) for the numerical coefficients in the expansion of \((a+b)^n\text{.}\) These binomial coefficients, which are given by Pascal’s triangle, can also be expressed using factorial notation as follows.

Binomial Coefficients.

For \(n \ge 0\) and \(0 \le k \le n\text{,}\)
\begin{equation*} ~_nC_k~ = \dfrac{n!}{(n-k)!~k!} \end{equation*}

Example 9.83.

Evaluate each binomial coefficient.
  1. \(\displaystyle ~_6C_2~\)
  2. \(\displaystyle ~_9C_8~\)
Solution.
  1. \(\displaystyle ~_6C_2~ = \dfrac{6!}{(6-2)!~2!} = \dfrac{6!}{4!~2!} = \dfrac{6 \cdot 5 \cdot 4!}{4!~\cdot 2 \cdot 1} = 15\)
  2. \(\displaystyle ~_9C_8~ = \dfrac{9!}{(9-8)!~8!} = \dfrac{9!}{1!~8!} = \dfrac{9 \cdot 8!}{1 \cdot 8!} = 9\)

Checkpoint 9.84. Practice 7.

Evaluate the binomial coefficient \(\quad_8C_6=\)
Answer.
\(28\)
Solution.
\(10\)

Note 9.85.

You can check that the formula works for \(k=0\) when we use \(0! = 1\text{.}\) You should find that
\begin{equation*} ~_nC_0~ = \dfrac{n!}{(n-0)!~0!} = \dfrac{n!}{n!~(1)} = 1 \end{equation*}
so the formula correctly gives us the coefficient of \(a^n\) in the expansion of \((a+b)^n\text{.}\)
We now have two methods for computing the binomial coefficient \(~_nC_k~\) : Pascal’s triangle and our formula using factorials. When \(n\) is small, especially if all the coefficients of the binomial expansion are needed, Pascal’s triangle is often the easier choice. But when \(n\) is large or if only one coefficient is required, the factorial formula for \(~_nC_k~\) is probably quicker than Pascal’s triangle.

Example 9.86.

Find the coefficient of \(m^{11}n^3\) in the expansion of \((m+n)^{14}\text{.}\)
Solution.
We are simply replacing \(a\) with \(m\) and \(b\) with \(n\) in the expansion of \((a+b)^{14}\text{,}\) so the term containing \(m^{11}n^3\) has \(n=14\) and \(k=3\text{,}\) and its coefficient is \(~_{14}C_3~\text{.}\)
\begin{equation*} ~_{14}C_3~ = \dfrac{14!}{(14-3)!~3!} = \dfrac{14 \cdot 13 \cdot 12 \cdot 11!}{11!~3!} = \dfrac{14 \cdot 13 \cdot 12}{3 \cdot 2 \cdot 1} = 364 \end{equation*}
The coefficient of \(m^{11}n^3\) is \(364\text{.}\)

Checkpoint 9.87. Exercise 7.

Find the coefficient of \(s^3t^4\) in the expansion of \((s+t)^7\text{.}\)
Answer:
Answer.
\(35\)
Solution.
\(35\)

Subsection The Binomial Theorem

Consider the expansion of \((a+b)^6\) written with the \(~_nC_k~\) notation for the coefficients.
\begin{align*} (a+b)^6 = \amp ~a^6b^0 + ~_6C_1~ a^5b^1 +~_6C_2~ a^4b^2 + ~_6C_3~ a^3b^3 + ~_6C_4~ a^2b^4 + ~_6C_5~ a^1b^5 + ~_6C_6~ a^0b^6\\ \amp k = 0~~~~~~~~k = 1~~~~~~~~~~~k = 2~~~~~~~~~~~~ k = 3~~~~~~~~~~~~ k = 4~~~~~~~~~~~~ k = 5~~~~~~~~~~~~ k = 6 \end{align*}
Check that each term can be written in the form
\begin{gather*} {}_6C_k~ a^{6-k}b^k \end{gather*}
for \(k=0\) to \(k=6\text{.}\) This means that we can use sigma notation to write the sum of the terms in the expansion.
\begin{gather*} (a+b)^6 = \sum_{k=0}^{6} ~_6C_k~ a^{6-k}b^k \end{gather*}
Notice that the sigma notation indicates terms for \(k=0\) to \(k=6\text{,}\) so there are 7 terms in the expansion, as there should be. Also note that the sum of the exponents on \(a\) and \(b\) is \((6-k)+k = 6\text{.}\)
The equation above is a special case of the binomial theorem, which uses everything we have learned to write the expanded form of a power of a binomial in the most compact form possible. We can write the general form of the theorem, for positive integers \(n\text{,}\) as follows.

The Binomial Theorem.

\begin{equation*} (a+b)^n = ~~\displaystyle{\sum_{k=0}^{n} ~_nC_k~ a^{n-k}b^k} \end{equation*}

Example 9.88.

  1. Use sigma notation to write the expanded form for \((x-2y)^{10}\text{.}\)
  2. Find the term containing \(y^7\text{,}\) and simplify.
Solution.
  1. We apply the binomial theorem with \(n=10\text{,}\) replacing \(a\) by \(x\) and \(b\) by \(-2y\text{,}\) to get
    \begin{equation*} (x-2y)^{10} = ~~\displaystyle{\sum_{k=0}^{10} ~_{10}C_k~ x^{10-k}(-2y)^k} \end{equation*}
  2. The term containing \(y^7\) corresponds to \(k=7\text{.}\) The \(k=7\) term looks like
    \begin{equation*} ~_{10}C_7~ x^{10-7}(-2y)^7~~~~\text{or} ~~~~ ~_{10}C_7~ x^3(-2y)^7 \end{equation*}
    The value of \(~_{10}C_7~\) is
    \begin{equation*} ~_{10}C_7~ = \dfrac{10!}{(10-7)!~7!} = \dfrac{10 \cdot 9 \cdot 8 \cdot 7!}{3!~7!} = \dfrac{10 \cdot 9 \cdot 8}{3 \cdot 2 \cdot 1} = 120 \end{equation*}
    and \((-2y)^7 = -128y^7\text{.}\) Thus, the term we want is
    \begin{equation*} ~_{10}C_7~ x^3 (-2y)^7 = 120x^3(-128y^7) = -15,360x^3y^7 \end{equation*}

Checkpoint 9.89. Practice 8.

  1. Use sigma notation to write the expanded form for \((2m-3n)^7\text{.}\)
    • \(\displaystyle {\displaystyle \sum_{k=1}^{7} \quad_7C_k (2m)^{7-k}(-3n)^{k} } \)
    • \(\displaystyle {\displaystyle \sum_{k=0}^{7} \quad_7C_k (2m)^{7-k}(-3n)^{k} } \)
    • \(\displaystyle {\displaystyle \sum_{k=0}^{7} \quad_nC_k (2m)^{n-k}(-3n)^{k} } \)
  2. Find the term containing \(m^3\text{,}\) and simplify.
Answer 1.
\(\text{Choice 2}\)
Answer 2.
\(22680m^{3}n^{4}\)
Solution.
  1. \(\displaystyle (2m-3n)^7 = \displaystyle{\sum_{k=0}^7 ~_7C_k~ (2m)^{7-k}(-3n)^k}\)
  2. \(\displaystyle ~_7C_4~ (2m)^3(-3n)^4 = 22,680m^3n^4\)

Subsection Section Summary

Subsubsection Vocabulary

  • Binomial coefficient
  • Pascal’s triangle
  • Factorial

Subsubsection CONCEPTS

  1. The binomial coefficient \(~_nC_k~ \) is the coefficient of the term containing \(b^k\) in the expansion of \((a+b)^n\text{.}\)
  2. The numbers in Pascal’s triangle are the binomial coefficients. Specifically, the number in the \(k^{\text{th}}\) position (starting with \(k=0\)) of the \(n^{\text{th}}\) row of the triangle is \(~_nC_k~ \text{.}\)
  3. If \(n\) is a positive integer,
    \begin{equation*} n! = n \cdot (n-1) \cdot (n-2) \cdot \cdots \cdot 3\cdot 2 \cdot 1 \end{equation*}
  4. For \(n \ge 0\) and \(0 \le k \le n\text{,}\)
    \begin{equation*} ~_nC_k~ = \dfrac{n!}{(n-k)!~k!} \end{equation*}
  5. The Binomial Theorem
    \begin{equation*} (a+b)^n = ~~\displaystyle{\sum_{k=0}^{n} ~_nC_k~ a^{n-k}b^k} \end{equation*}

Subsubsection STUDY QUESTIONS

  1. What does the symbol \(~_nC_k~ \) denote?
  2. Explain how to use Pascal’s triangle when expanding a power of a binomial.
  3. What is factorial notation?
  4. State a formula for the binomial coefficients in factorial notation.
  5. Explain how to use the binomial theorem to expand \((a+b)^7\text{.}\)

Subsubsection SKILLS

Practice each skill in the Homework Problems listed.
  1. Expand powers of binomials: #1–22, 33–34
  2. Simplify factorial notation: #23–24
  3. Evaluate binomial coefficients: #25–26
  4. Find specific terms of binomial expansions: #28–32, #35-42
  5. Use Pascal’s triangle: #43–44

Exercises Homework 9.5

Exercise Group.

For Problems 1–4, answer the questions without expanding the power.
1.
How many terms are in the expansion of \((a+b)^{50}\text{?}\) In the expansion of \((2x+3y)^{100}\text{?}\)
2.
How many terms are in the expansion of \((a+b)^{75}\text{?}\) In the expansion of \((5x-7y)^{200}\text{?}\)
3.
What is the sum of the exponents on \(x\) and \(y\) in each term of the expansion of \((x+y)^{100}\text{?}\) In the expansion of \((8x-7y)^{50}\text{?}\)
4.
What is the sum of the exponents on \(x\) and \(y\) in each term of the expansion of \((x+y)^{200}\text{?}\) In the expansion of \((9x+4y)^{75}\text{?}\)

5.

Write down the portion of Pascal’s triangle corresponding to rows from \(n=0\) to \(n=10\text{.}\) How many rows are involved?

6.

Write down the portion of Pascal’s triangle corresponding to rows from \(n=0\) to \(n=12\text{.}\) How many rows are involved?

Exercise Group.

For Problems 7–18, write the power in expanded form.
7.
\((x+3)^5\)
8.
\((2x+y)^4\)
9.
\((z-3)^4\)
10.
\((2w-1)^5\)
11.
\(\left(2x-\dfrac{y}{2}\right)^3\)
12.
\(\left(\dfrac{x}{3}+3\right)^6\)
13.
\((x^2-3)^7\)
14.
\(\left(1-y^2\right)^5\)
15.
\((x+y)^5\)
16.
\((x-y)^6\)
17.
\((p-2q)^4\)
18.
\((m+3n)^8\)

19.

Simplify \((1+5t)^3 + (1-5t)^3\text{.}\)

20.

Simplify \((3+2a)^4 + (3-2a)^4.\)

21.

Write \(\left(z-\dfrac{1}{z}\right)^5\) in descending powers of \(z\text{.}\)

22.

Write \(\left(v+\dfrac{1}{v}\right)^4\) in descending powers of \(v\text{.}\)

Exercise Group.

Write the expressions in Problems 23–24 in expanded form and simplify.
23.
  1. \(\displaystyle 5!\)
  2. \(\displaystyle \dfrac{9!}{7!}\)
  3. \(\displaystyle \dfrac{5!~7!}{12!}\)
  4. \(\displaystyle \dfrac{8!}{2!(8-2)!}\)
24.
  1. \(\displaystyle 7!\)
  2. \(\displaystyle \dfrac{12!}{11!}\)
  3. \(\displaystyle \dfrac{12!~4!}{16!}\)
  4. \(\displaystyle \dfrac{10!}{4!(10-4)!}\)

Exercise Group.

In Problems 25–26, evaluate the binomial coefficient.
25.
  1. \(\displaystyle ~_9C_6~ \)
  2. \(\displaystyle ~_{12}C_3~ \)
  3. \(\displaystyle ~_{20}C_{18}~ \)
  4. \(\displaystyle ~_{14}C_9~ \)
26.
  1. \(\displaystyle ~_8C_5~ \)
  2. \(\displaystyle ~_{13}C_4~ \)
  3. \(\displaystyle ~_{18}C_{16}~ \)
  4. \(\displaystyle ~_{16}C_7~ \)

27.

Find in ascending powers of \(x\) the first three terms in the expansions of
  1. \(\displaystyle (1-2x)^7\)
  2. \(\displaystyle (2-x)^6\)

28.

Find in descending powers of \(y\) the first three terms in the expansions of
  1. \(\displaystyle (5y-2)^4\)
  2. \(\displaystyle (3y+1)^5\)

29.

Find in descending powers of \(u\) the first three non-zero terms of \((4u-1)^5-(4u+1)^5\text{.}\)

30.

Find in ascending powers of \(t\) the first three non-zero terms of \((1-3t)^6-(1+3t)^6\text{.}\)

31.

Find the first four terms of
  1. \(\displaystyle (1-5c)^6\)
  2. \(\displaystyle (1-4c)(1-5c)^6\)

32.

Find the first four terms of
  1. \(\displaystyle (1+2a)^9\)
  2. \(\displaystyle (1+3a)(1+2a)^9\)

33.

Simplify \(\left(1+\sqrt{3}\right)^4 + \left(1-\sqrt{3}\right)^4\text{.}\)

34.

Simplify \(\left(\sqrt{2}+\sqrt{3}\right)^4 + \left(\sqrt{2}-\sqrt{3}\right)^4\text{.}\)

Exercise Group.

For Problems 35–42, find the coefficient of the indicated term.
35.
\((x+y)^{20};~~ x^{13}y^7\)
36.
\((x+y)^{15};~~ x^{12}y^3\)
37.
\((a-2b)^{12};~~ a^5b^7\)
38.
\((2a-b)^{12};~~ a^8b^4\)
39.
\(\left(x-\sqrt{2}\right)^{10};~~ x^4\)
40.
\(\left(x-\dfrac{1}{2}\right)^8;~~ x^5\)
41.
\(\left(a^3-b^3\right)^9;~~ a^{18}b^9\)
42.
\(\left(x^2-y^2\right)^7;~~ x^{10}y^4 \)

Exercise Group.

For Problems 43–44, refer to Pascal’s triangle to answer the questions.
43.
Write out the terms of the sequence \(11^n\) for \(n = 0, 1, 2, 3, 4\text{.}\) How are these related to the rows of Pascal’s triangle? Can you explain why? (Hint: Rewrite \(11\) as \(10+1\text{.}\))
44.
Write out the terms of the sequence \(1.1^n\) for \(n = 0, 1, 2, 3, 4\text{.}\) How are these related to the rows of Pascal’s triangle? Can you explain why? (Hint: Rewrite \(1.1\) as \(1+0.1\text{.}\))
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