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Binomial expansion and Pascal's triangle

Learning goals

  • Uses Pascal's triangle to obtain binomial coefficients.
  • Finds the terms of a binomial power using its binomial coefficients.
  • Determines specific coefficients or terms without fully expanding when possible.
  • Connects coefficients to combinatorial structure.

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Binomial expansion and Pascal's triangle 1
Write the row of Pascal's triangle that gives the coefficients for expanding \((a+b)^4\).

Hints

- The row begins and ends with \(1\). - Each interior entry is the sum of the two entries above it in Pascal's triangle. - A fourth power has five coefficient entries.

Solution

1. The coefficient row for the fourth power is \(1,4,6,4,1\). 2. These are the coefficients that multiply the five terms in the expansion of \((a+b)^4\).

Answer

\(1,4,6,4,1\)
Binomial expansion and Pascal's triangle 2
Pascal's triangle is numbered so that the top row is row \(n=0\). Which row supplies the coefficients for expanding \((a+b)^7\), and how many coefficients are in that row?

Hints

- Compare the binomial exponent with the row-number convention stated in the problem. - Think about how many terms appear in a binomial expansion of exponent \(n\).

Solution

1. The exponent of a binomial power matches the Pascal row number when the top row is \(n=0\). 2. Therefore, \((a+b)^7\) uses row \(n=7\). 3. A row for exponent \(n\) contains \(n+1\) coefficients, so row \(7\) contains \(8\) coefficients.

Answer

Row \(n=7\), with \(8\) coefficients.
Binomial expansion and Pascal's triangle 3
A row of Pascal's triangle is \(1,\ 2,\ 1\). Write the next row. Briefly explain how you found its middle numbers.

Hints

- How do the outside numbers change from one Pascal row to the next? - Each interior number comes from a pair of neighbors in the previous row.

Solution

1. Keep \(1\) at each end of the next row. 2. Add neighboring numbers: \(1+2=3\) and \(2+1=3\). 3. The next row is \(1,\ 3,\ 3,\ 1\).

Answer

\(1,\ 3,\ 3,\ 1\). Each interior number is the sum of the two numbers above it.
Binomial expansion and Pascal's triangle 4
Row \(4\) of Pascal's triangle is \(1,\ 4,\ 6,\ 4,\ 1\). What is the coefficient of \(x^2y^2\) in \((x+y)^4\)?

Hints

- Match the Pascal row entries to the expansion terms in order. - Which term of a fourth-power expansion uses two factors of each variable?

Solution

1. The coefficients in the expansion of \((x+y)^4\) follow row \(4\) of Pascal's triangle. 2. The term \(x^2y^2\) is the middle term, with coefficient \(6\).

Answer

\(6\)
Binomial expansion and Pascal's triangle 5
A student claims that \(x^3y^2\) appears when \((x+y)^4\) is expanded. Could that term appear? If not, what power of \(y\) must go with \(x^3\)? Explain by thinking about the four factors.

Hints

- How many separate parentheses are multiplied in a fourth power? - Each term takes one choice from each parenthesis. Can the proposed term come from only four choices?

Solution

1. Every term comes from choosing one letter from each of four factors \((x+y)\). 2. The proposed term \(x^3y^2\) requires five choices, so it cannot appear. 3. Choosing \(x\) from three factors leaves one factor contributing \(y\), so the term uses \(y^1\).

Answer

No. Each term chooses one letter from each of four factors. Three choices of \(x\) leave only one choice of \(y\), so the term is \(x^3y\), not \(x^3y^2\).
Binomial expansion and Pascal's triangle 6
When you multiply \((a+b)(a+b)\), which products contribute to \(ab\)? What is the coefficient of \(ab\) after combining them? Explain.

Hints

- Choose one term from each parenthesis. Which choices give one \(a\) and one \(b\)? - Do two products with the same letters in a different order combine?

Solution

1. Choosing \(a\) then \(b\) gives the product \(ab\). 2. Choosing \(b\) then \(a\) also gives \(ab\). 3. These two contributions combine to \(2ab\), so the coefficient is \(2\).

Answer

Choosing \(a\) then \(b\) and choosing \(b\) then \(a\) each give \(ab\). Together they give \(2ab\), so the coefficient is \(2\).
Binomial expansion and Pascal's triangle 7
Complete the expansion \((x-3)^2=x^2+\square x+9\). Find the missing coefficient.

Hints

- Write the square as a product of two identical factors. - Which two products give the terms containing one \(x\)? - Keep the negative sign when combining them.

Solution

1. Expand \((x-3)(x-3)\) to get \(x^2-3x-3x+9\). 2. Combine the linear terms: \(-3x-3x=-6x\). 3. The missing coefficient is \(-6\).

Answer

\(-6\)
Binomial expansion and Pascal's triangle 8
In Pascal's triangle, every interior entry is the sum of the two entries above it. a) In row \(n=13\), \(\binom{13}{6}=1716\) and \(\binom{13}{7}=1716\). Use Pascal's rule to find \(\binom{14}{7}\). b) Use symmetry to explain why \(\binom{13}{6}\) and \(\binom{13}{7}\) are equal. c) State \(\binom{15}{0}\) and \(\binom{15}{15}\) without writing all rows through \(n=15\).

Hints

- How is an entry formed from the two entries above it? - Compare the left and right sides of a row. - What pattern appears at the two ends of every row?

Solution

1. Pascal's rule gives \(\binom{14}{7}=\binom{13}{6}+\binom{13}{7}=1716+1716=3432\). 2. The symmetry identity is \(\binom{n}{k}=\binom{n}{n-k}\). Since \(13-6=7\), \(\binom{13}{6}=\binom{13}{7}\). 3. Every row begins and ends with \(1\), so \(\binom{15}{0}=1\) and \(\binom{15}{15}=1\).

Answer

a) \(\binom{14}{7}=3432\) b) Since \(13-6=7\), symmetry gives \(\binom{13}{6}=\binom{13}{7}\). c) \(\binom{15}{0}=1\), \(\binom{15}{15}=1\)
Binomial expansion and Pascal's triangle 9
Each row of Pascal's triangle is symmetric. a) In row \(n=7\), the number \(35\) appears twice. Write \(35\) in two ways as binomial coefficients of the form \(\binom{n}{k}\). b) Compute \(\binom{10}{8}\). Explain how the symmetry identity \(\binom{n}{k}=\binom{n}{n-k}\) simplifies the calculation.

Hints

- Recall the entries and indexing of row \(n=7\). - Which two values of \(k\) are symmetric in a row? - Replace a large \(k\) with \(n-k\) before calculating. - Compare the number of factors needed for \(\binom{10}{8}\) and \(\binom{10}{2}\).

Solution

1. Row \(n=7\) contains \(1,7,21,35,35,21,7,1\). Thus \(35=\binom{7}{3}=\binom{7}{4}\). 2. By symmetry, \(\binom{10}{8}=\binom{10}{2}\). 3. \(\binom{10}{2}=\frac{10\cdot 9}{2}=45\). Using \(k=2\) requires fewer factors than calculating directly with \(k=8\).

Answer

a) \(\binom{7}{3}\) and \(\binom{7}{4}\) b) \(\binom{10}{8}=\binom{10}{2}=45\)
Binomial expansion and Pascal's triangle 10
A fair coin is flipped five times. The row of Pascal's triangle for four flips is \(1,4,6,4,1\). a) Write the next row of Pascal's triangle. b) How many head-tail sequences have exactly three heads? c) How many have four or five heads? d) What is the probability of getting exactly one head?

Hints

- How is a new Pascal row built from the previous one? - Match each entry with the number of heads, starting at zero. - Count the sequences with four and five heads separately, then combine them. - For a fair coin, are all five-flip sequences equally likely?

Solution

1. Add neighboring entries of the four-flip row, keeping \(1\) at each end. The five-flip row is \(1,5,10,10,5,1\). 2. The entry for exactly three heads is \(\binom{5}{3}=10\). 3. Four or five heads give \(\binom{5}{4}+\binom{5}{5}=5+1=6\) sequences. 4. Exactly one head occurs in \(\binom{5}{1}=5\) of the \(2^5=32\) equally likely sequences. Its probability is \(\frac{5}{32}\).

Answer

a) \(1,5,10,10,5,1\) b) \(10\) c) \(6\) d) \(\frac{5}{32}\)
Binomial expansion and Pascal's triangle 11
The polynomial function \(f\) is defined by \(f(x) = x^3 + 3x^2 + 3x + 1\). 1. Use a binomial identity to write \(f(x)\) in the form \((x + a)^n\). 2. Find the zero of \(f\) and state its multiplicity.

Hints

- Compare the coefficients with the expansion of \((a + b)^3\). - A power of a linear factor equals zero only when the linear factor equals zero. - The exponent on the factor gives the multiplicity.

Solution

1. The identity \((a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) with \(a = x\) and \(b = 1\) gives \(f(x) = (x + 1)^3\). 2. Set \((x + 1)^3 = 0\). Then \(x + 1 = 0\), so \(x = -1\). 3. Because the factor \(x + 1\) occurs to the third power, the zero has multiplicity 3.

Answer

1. \(f(x) = (x + 1)^3\) 2. \(x = -1\) with multiplicity 3
Binomial expansion and Pascal's triangle 12
Each branch of the tree represents a choice when multiplying \((a+b)^3\). a) List the three choices along each path ending at P. b) Which term do these paths all produce? What is its coefficient in \((a+b)^3\)? Explain using the tree.
Figure for problem 537456

Hints

- To avoid missing a path, finish exploring one branch before moving to the next. - Does changing the order of factors change their product? - How could several copies of the same product be written as one term?

Solution

1. The marked paths are \(a\to a\to b\), \(a\to b\to a\), and \(b\to a\to a\). 2. Each path multiplies two \(a\) factors and one \(b\), giving \(a^2b\). 3. Three paths give the same term, so their contributions add to \(3a^2b\). The coefficient is \(3=\binom{3}{2}\).

Answer

a) \(a\to a\to b\), \(a\to b\to a\), and \(b\to a\to a\) b) All give \(a^2b\); the coefficient is \(3\) because there are three such paths.
Binomial expansion and Pascal's triangle 13
Use Pascal's triangle to expand \((x+y)^4\).

Hints

- Use the Pascal row associated with a fourth power. - The exponent of \(x\) decreases from \(4\) to \(0\) while the exponent of \(y\) increases from \(0\) to \(4\). - Each term has total exponent \(4\).

Solution

1. The coefficient row for the fourth power is \(1,4,6,4,1\). 2. Pair these coefficients with descending powers of \(x\) and ascending powers of \(y\). 3. The expansion is \(x^4+4x^3y+6x^2y^2+4xy^3+y^4\).

Answer

\(x^4+4x^3y+6x^2y^2+4xy^3+y^4\)
Binomial expansion and Pascal's triangle 14
Use Pascal's triangle to expand \((x-y)^4\).

Hints

- Start with the same coefficient row as for \((x+y)^4\). - Remember that the second binomial term is \(-y\), not \(y\). - Track whether each power of \(-y\) is positive or negative.

Solution

1. Use coefficients \(1,4,6,4,1\). 2. Treat the second term as \(-y\). Its powers alternate the signs of successive terms. 3. The expansion is \(x^4-4x^3y+6x^2y^2-4xy^3+y^4\).

Answer

\(x^4-4x^3y+6x^2y^2-4xy^3+y^4\)
Binomial expansion and Pascal's triangle 15
Find the coefficient of \(x^3y^2\) in the expansion of \((x+2y)^5\) without expanding the entire binomial.

Hints

- Determine how many times the second binomial term must be selected to produce \(y^2\). - Use a binomial coefficient to count the choices of those factor positions. - Include the numerical coefficient carried by each selected \(2y\) factor.

Solution

1. To obtain \(x^3y^2\), choose the \(2y\) term from exactly \(2\) of the \(5\) factors. 2. The corresponding term is \(\binom{5}{2}x^3(2y)^2\). 3. Its coefficient is \(10\cdot4=40\).

Answer

\(40\)
Binomial expansion and Pascal's triangle 16
Maarja writes \((a-b)^4=a^4-4a^3b-6a^2b^2-4ab^3+b^4\). Identify Maarja's error and write the correct expansion.

Hints

- Keep the second binomial term as \(-b\) while considering its powers. - Compare what happens to even and odd powers of a negative quantity. - The Pascal coefficients themselves do not carry the alternating signs.

Solution

1. The coefficients \(1,4,6,4,1\) are correct, but the signs must come from powers of \(-b\). 2. Even powers of \(-b\) are positive and odd powers are negative. 3. Therefore, the middle term \(6a^2b^2\) must be positive. 4. The correct expansion is \(a^4-4a^3b+6a^2b^2-4ab^3+b^4\).

Answer

The \(a^2b^2\) term has the wrong sign. The correct expansion is \(a^4-4a^3b+6a^2b^2-4ab^3+b^4\).
Binomial expansion and Pascal's triangle 17
In the expansion of \((a+b)^5\), explain why the exponents on \(a\) and \(b\) add to \(5\) in every term and why the coefficient sequence is symmetric.

Hints

- Think of the expansion as making one choice from each of five factors. - Count how many factors contribute \(a\) and how many contribute \(b\). - Compare choosing positions for \(b\) with choosing the complementary positions for \(a\).

Solution

1. Each term in the expansion comes from choosing either \(a\) or \(b\) from each of the \(5\) identical factors. 2. If \(b\) is chosen from \(k\) factors, then \(a\) is chosen from the remaining \(5-k\) factors. The term therefore has exponents whose sum is \((5-k)+k=5\). 3. The coefficient is \(\binom{5}{k}\). Choosing the \(k\) positions for \(b\) is equivalent to choosing the \(5-k\) positions for \(a\), so \(\binom{5}{k}=\binom{5}{5-k}\). This gives symmetric coefficients.

Answer

Each of the \(5\) factors contributes either one \(a\) or one \(b\), so the exponents total \(5\). The coefficients are symmetric because \(\binom{5}{k}=\binom{5}{5-k}\).
Binomial expansion and Pascal's triangle 18
Row \(4\) of Pascal's triangle is \(1,\ 4,\ 6,\ 4,\ 1\). A student writes the next row as \(1,\ 5,\ 11,\ 11,\ 5,\ 1\). Find the incorrect entries using the addition rule, and write the correct row.

Hints

- Each number inside a Pascal row comes from two numbers in the row above. - Check all the interior entries, even when the new row looks symmetric. - Which additions using row \(4\) should give the entries of the next row?

Solution

1. Each interior entry is the sum of the two entries above it. Both middle entries come from \(4+6=10\) or \(6+4=10\), not \(11\). 2. Replace both \(11\)'s with \(10\)'s: \(1,\ 5,\ 10,\ 10,\ 5,\ 1\).

Answer

Both \(11\)'s should be \(10\)'s, since \(4+6=10\). Correct row: \(1,\ 5,\ 10,\ 10,\ 5,\ 1\).
Binomial expansion and Pascal's triangle 19
Without expanding every term, find the coefficient of \(x^2\) in \((x+3)^3\).

Hints

- To get \(x^2\), how many factors contribute \(x\)? - Which factor supplies the remaining constant? - Count how many positions that constant could occupy.

Solution

1. The \(x^2\) term comes from choosing \(x\) twice and \(3\) once from the three factors. 2. There are \(3\) ways to choose which factor contributes \(3\). 3. The coefficient is \(3\cdot3=9\).

Answer

\(9\)
Binomial expansion and Pascal's triangle 20
Complete the two missing coefficients in \((2x-y)^3=8x^3+\square x^2y+\square xy^2-y^3\).

Hints

- Which coefficients appear in the third row of Pascal's triangle? - How do the exponents of \(2x\) and \(-y\) change from term to term? - Which of the missing terms has an odd power of the negative quantity?

Solution

1. The cube uses binomial coefficients \(1,3,3,1\) with alternating signs because the second term is \(-y\). 2. The \(x^2y\) term is \(3(2x)^2(-y)=-12x^2y\). 3. The \(xy^2\) term is \(3(2x)(-y)^2=6xy^2\).

Answer

\(-12\) and \(6\), in that order.
Binomial expansion and Pascal's triangle 21
Three lights are arranged in a row. Each light can glow red or blue. The expansion of \((r+b)^3\) can count the color sequences. Expand \((r+b)^3\). Which coefficient counts the left-to-right sequences with exactly one red light? List those sequences to check the count, using R for red and B for blue.

Hints

- Which Pascal row gives the coefficients of a binomial cube? - Which term uses one \(r\) and two \(b\)'s? - Check the coefficient by listing the positions of the red light.

Solution

1. Expand using coefficients \(1,3,3,1\): \((r+b)^3=r^3+3r^2b+3rb^2+b^3\). 2. Exactly one red light means one \(r\) and two \(b\)'s. The coefficient of \(rb^2\) is \(3\). 3. The sequences are \(\text{RBB}\), \(\text{BRB}\), and \(\text{BBR}\), confirming the coefficient.

Answer

\((r+b)^3=r^3+3r^2b+3rb^2+b^3\). The coefficient of \(rb^2\) is \(3\); the sequences are \(\text{RBB}\), \(\text{BRB}\), and \(\text{BBR}\).
Binomial expansion and Pascal's triangle 22
Pascal's triangle is closely connected to binomial coefficients. a) One row begins \(1,9,36,84,126,126,\ldots\). Find the first six entries of the next row. b) In row \(n=10\), \(\binom{10}{3}=120\). Find \(\binom{10}{4}\) using \(\binom{n}{k+1}=\binom{n}{k}\frac{n-k}{k+1}\) or the binomial coefficient formula. c) The sum of the entries in row \(n\) is \(2^n\). Verify this for row \(n=5\).

Hints

- Use the local relationship between neighboring entries in Pascal's triangle. - Relate adjacent binomial coefficients in the same row without recomputing the entire row. - For the row sum, compare the direct total with the stated power-of-two pattern.

Solution

1. Add adjacent entries from the given row: the next row begins \(1\), \(1+9=10\), \(9+36=45\), \(36+84=120\), \(84+126=210\), and \(126+126=252\). 2. \(\binom{10}{4}=120\cdot\frac{10-3}{4}=120\cdot\frac{7}{4}=210\). 3. Row \(n=5\) is \(1,5,10,10,5,1\). Its sum is \(1+5+10+10+5+1=32=2^5\).

Answer

a) \(1,10,45,120,210,252\) b) \(\binom{10}{4}=210\) c) \(1+5+10+10+5+1=32=2^5\)
Binomial expansion and Pascal's triangle 23
A six-stage experiment has two equally likely outcomes at each stage. The number of paths with \(k\) successes is the coefficient of the corresponding term in \((a+b)^6\). a) Give the coefficients in row \(n=6\) of Pascal's triangle. b) A multiple-choice quiz has \(6\) questions, each with two choices and exactly one correct answer. How many answer patterns contain exactly \(3\) correct answers? c) Sofia guesses randomly on all \(6\) questions. Find the probability of at least \(5\) correct answers.

Hints

- Form the next Pascal's triangle row by adding adjacent entries. - Match each coefficient with its number of correct answers. - “At least \(5\)” includes \(5\) and \(6\). - How many total answer patterns are possible for \(6\) two-choice questions?

Solution

1. Row \(n=6\) is \(1,6,15,20,15,6,1\). 2. Exactly \(3\) correct answers can occur in \(\binom{6}{3}=20\) answer patterns. 3. At least \(5\) correct answers means \(5\) or \(6\). The number of favorable patterns is \(\binom{6}{5}+\binom{6}{6}=6+1=7\). There are \(2^6=64\) equally likely patterns, so the probability is \(\frac{7}{64}=0.109375\).

Answer

a) \(1,6,15,20,15,6,1\) b) \(20\) c) \(\frac{7}{64}=0.109375\)
Binomial expansion and Pascal's triangle 24
Compare these functions: \(f_1(x)=(x-3)^3\) \(f_2(x)=x^3-27\) \(f_3(x)=(3-x)^3\) \(f_4(x)=(x-3)(x^2+3x+9)\) \(f_5(x)=x^3-9x^2+27x-27\). a) Evaluate every function at \(x=3\) and \(x=0\). b) Which functions are identical? Justify by expanding or using an identity. Explain why matching values at the two inputs in part a does not, by itself, prove that two functions are identical.

Hints

- Compare the signs of \((x-3)^3\) and \((3-x)^3\). - An expanded cube and a difference of cubes give two different useful identities. - Could two different polynomials agree at a few points but not at every input?

Solution

1. All five values at \(x=3\) are \(0\). At \(x=0\), the values of \(f_1,f_2,f_3,f_4,f_5\) are \(-27,-27,27,-27,-27\), respectively. 2. The cube identity gives \((x-3)^3=x^3-9x^2+27x-27\), so \(f_1=f_5\). 3. The difference-of-cubes identity gives \(x^3-27=(x-3)(x^2+3x+9)\), so \(f_2=f_4\). 4. Because \((3-x)^3=-(x-3)^3\), function \(f_3\) is not identical to any other listed function. Matching at two points is not enough: \(f_1(1)=-8\) but \(f_2(1)=-26\), even though both agree at \(0\) and \(3\).

Answer

a) At \(x=3\), all values are \(0\). At \(x=0\): \(f_1=-27\), \(f_2=-27\), \(f_3=27\), \(f_4=-27\), \(f_5=-27\). b) \(f_1=f_5\) because \((x-3)^3=x^3-9x^2+27x-27\). Also \(f_2=f_4\) because \(x^3-27=(x-3)(x^2+3x+9)\). Matching at two inputs does not prove identity: \(f_1(1)=-8\) but \(f_2(1)=-26\).
Binomial expansion and Pascal's triangle 25
A basketball player makes a free throw with probability \(p=0.8\). The player takes three free throws. a) Find the probability of making exactly two shots. b) Compare your calculation with the expansion of \((p+q)^3\), where \(q=1-p\) is the probability of a miss. Which term in the expansion represents exactly two made shots?
Figure for problem 537455

Hints

- List the sequences with exactly two makes and one miss. - Count how many such sequences there are. - Compare that count with the coefficients in the expansion of \((p+q)^3\).

Solution

1. The probability of a make is \(p=0.8\), and the probability of a miss is \(q=1-0.8=0.2\). 2. Exactly two makes occur in the sequences MMX, MXM, and XMM, where M represents a make and X represents a miss. 3. Each sequence has probability \(p^2q=(0.8)^2\cdot 0.2=0.128\). Therefore, \(P(\text{exactly two makes})=3p^2q=3\cdot(0.8)^2\cdot 0.2=0.384\). 4. The binomial expansion is \((p+q)^3=p^3+3p^2q+3pq^2+q^3\). The term \(3p^2q\) represents exactly two made shots and one miss.

Answer

a) \(0.384=38.4\%\) b) The term is \(3p^2q\).
Binomial expansion and Pascal's triangle 26
Use binomial coefficients to expand \((2x+y)^4\).

Hints

- Use the fourth-power coefficient row, but do not forget that the first binomial term is \(2x\). - Apply each exponent to the entire term \(2x\). - Simplify numerical powers after setting up all five terms.

Solution

1. Use coefficients \(1,4,6,4,1\). 2. Substitute \(2x\) for the first binomial term: \((2x)^4+4(2x)^3y+6(2x)^2y^2+4(2x)y^3+y^4\). 3. Simplify to obtain \(16x^4+32x^3y+24x^2y^2+8xy^3+y^4\).

Answer

\(16x^4+32x^3y+24x^2y^2+8xy^3+y^4\)
Binomial expansion and Pascal's triangle 27
Expand \((x-2)^5\) using binomial coefficients.

Hints

- Use the Pascal coefficients for a fifth power. - Keep \(-2\) inside parentheses when raising it to powers. - Check that the signs alternate because the second binomial term is negative.

Solution

1. Use the fifth-power coefficients \(1,5,10,10,5,1\). 2. Treat the second term as \(-2\): \(x^5+5x^4(-2)+10x^3(-2)^2+10x^2(-2)^3+5x(-2)^4+(-2)^5\). 3. Simplify to get \(x^5-10x^4+40x^3-80x^2+80x-32\).

Answer

\(x^5-10x^4+40x^3-80x^2+80x-32\)
Binomial expansion and Pascal's triangle 28
The polynomial \(x^4-8x^3+24x^2-32x+16\) is the expansion of a fourth power of a binomial. Identify the binomial expression and explain how the coefficients and signs support your answer.

Hints

- Compare the coefficient pattern with a Pascal row for a fourth power. - The alternating signs suggest a negative second binomial term. - Use the constant term to infer the magnitude of that second term.

Solution

1. A fourth-power binomial uses Pascal coefficients \(1,4,6,4,1\). 2. The expression \((x-2)^4\) expands with terms \(x^4\), \(4x^3(-2)\), \(6x^2(-2)^2\), \(4x(-2)^3\), and \((-2)^4\). 3. These simplify to \(x^4-8x^3+24x^2-32x+16\), matching the given polynomial.

Answer

\((x-2)^4\)
Binomial expansion and Pascal's triangle 29
The tree shows the first two choices when the same binomial is multiplied four times. Each branch tells you which term is chosen from that binomial. a) Which binomial is being raised to the fourth power? b) For each path shown, how many ways can the last two choices give exactly two \(x\)'s in all four factors? Give the four counts from top to bottom. c) Find the coefficient of \(x^2\) in the expansion. Explain why the counts from part b give this coefficient.
Figure for problem 551157

Hints

- Read the two different branch labels; they are the terms of the binomial. - For each shown path, how many more \(x\) choices are needed to get exactly two? - A path contributes a product, not just one count. What do the other choices contribute?

Solution

1. Each branching offers \(x\) or \(2\), so the binomial power is \((x+2)^4\). 2. From top to bottom, the paths begin \(xx\), \(x2\), \(2x\), and \(22\). The valid ending counts for exactly two \(x\) choices are \(1,2,2,1\), a total of \(6\). 3. Each completed path with two \(x\)'s also has two factors of \(2\), giving \(4x^2\). The six paths contribute \(6\cdot4x^2=24x^2\), so the coefficient is \(24\).

Answer

a) \((x+2)^4\) b) \(1,2,2,1\), from top to bottom; \(6\) paths in total c) \(24\), because the six paths each contribute \(4x^2\).
Binomial expansion and Pascal's triangle 30
For an integer \(n\ge4\), consider \((a+b)^n\). a) Write its first three terms in descending powers of \(a\), without using sigma notation. b) Write the term containing \(a^{n-k}b^k\) and state the possible whole-number values of \(k\). c) Explain why its coefficient counts choices from the \(n\) factors.

Hints

- Think of multiplying \(n\) copies of \(a+b\); each term uses one choice from each factor. - For the first three terms, consider choosing \(b\) zero, one, or two times. - If exactly \(k\) factors supply \(b\), how many ways can you choose those factors?

Solution

1. Choosing \(b\) zero, one, or two times gives the first three terms: \(a^n+na^{n-1}b+\binom{n}{2}a^{n-2}b^2\). 2. Choosing \(b\) from exactly \(k\) factors gives \(\binom{n}{k}a^{n-k}b^k\), where \(k\) is a whole number from \(0\) through \(n\). 3. There are \(\binom{n}{k}\) ways to choose which \(k\) of the \(n\) factors supply \(b\). Each choice produces the same term, so the number of choices is its coefficient.

Answer

a) \(a^n+na^{n-1}b+\binom{n}{2}a^{n-2}b^2\) b) \(\binom{n}{k}a^{n-k}b^k\), for whole numbers \(0\le k\le n\) c) The coefficient counts the ways to choose the \(k\) factors that contribute \(b\).
Binomial expansion and Pascal's triangle 31
In the expansion of \((2x+k)^6\), the coefficient of \(x^3\) is \(4320\). Find \(k\), given that \(k>0\).

Hints

- First determine how many factors must contribute the \(2x\) term to produce \(x^3\). - Use a binomial coefficient for the number of position choices. - Include both the power of \(2\) and the power of \(k\) in the coefficient equation.

Solution

1. An \(x^3\) term occurs when \(2x\) is chosen from \(3\) of the \(6\) factors and \(k\) is chosen from the other \(3\). 2. The coefficient of \(x^3\) is \(\binom{6}{3}2^3k^3=20\cdot8\cdot k^3=160k^3\). 3. Set \(160k^3=4320\). Then \(k^3=27\), so \(k=3\) because \(k>0\).

Answer

\(k=3\)
Binomial expansion and Pascal's triangle 32
In the expansion of \((1+x)^n\), where \(n\) is a positive integer, the coefficient of \(x^2\) is \(45\). a) Find \(n\). b) Find the coefficient of \(x^3\) in the same expansion.

Hints

- Express the coefficient of \(x^2\) using a binomial coefficient involving the unknown exponent. - Rewrite that binomial coefficient as an algebraic expression in \(n\) and solve the resulting equation under the positive-integer condition. - Once \(n\) is known, use the corresponding binomial coefficient for the \(x^3\) term.

Solution

1. In \((1+x)^n\), the coefficient of \(x^2\) is \(\binom{n}{2}\). Therefore, \(\binom{n}{2}=45\). 2. Using \(\binom{n}{2}=\frac{n(n-1)}{2}\), solve \(\frac{n(n-1)}{2}=45\). This gives \(n^2-n-90=0\), so \((n-10)(n+9)=0\). 3. Since \(n\) is positive, \(n=10\). 4. The coefficient of \(x^3\) is then \(\binom{10}{3}=120\).

Answer

a) \(n=10\) b) \(120\)

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