IB Mathematics: Analysis and Approaches Quiz: Binomial Theorem
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Binomial TheoremQuestion 1 of 20

What is the ratio of the coefficient of x7x^7 to the coefficient of x6x^6 in the expansion of (2x+3)10(2x+3)^{10}?

47\frac{4}{7}
67\frac{6}{7}
218\frac{21}{8}
821\frac{8}{21}
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Binomial Theorem

Practice Binomial Theorem in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Binomial Theorem, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

What is the ratio of the coefficient of x7x^7 to the coefficient of x6x^6 in the expansion of (2x+3)10(2x+3)^{10}?

  1. 47\frac{4}{7}
  2. 67\frac{6}{7}
  3. 218\frac{21}{8}
  4. 821\frac{8}{21} (correct answer)
Explanation: When you encounter binomial expansion problems asking for ratios of coefficients, you're working with the binomial theorem: (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k For (2x+3)10(2x+3)^{10}, the general term is (10k)(2x)10−k(3)k=(10k)210−k3kx10−k\binom{10}{k}(2x)^{10-k}(3)^k = \binom{10}{k}2^{10-k}3^k x^{10-k} To find the coefficient of x7x^7, you need the term where 10−k=710-k = 7, so k=3k = 3: (103)2733=120⋅128⋅27=414,720\binom{10}{3}2^7 3^3 = 120 \cdot 128 \cdot 27 = 414,720 For the coefficient of x6x^6, you need 10−k=610-k = 6, so k=4k = 4: (104)2634=210⋅64⋅81=1,088,640\binom{10}{4}2^6 3^4 = 210 \cdot 64 \cdot 81 = 1,088,640 The ratio is: 414,7201,088,640=821\frac{414,720}{1,088,640} = \frac{8}{21} This confirms answer D is correct. Answer A (47\frac{4}{7}) likely comes from incorrectly using just the binomial coefficients (103)(104)=120210=47\frac{\binom{10}{3}}{\binom{10}{4}} = \frac{120}{210} = \frac{4}{7}, ignoring the powers of 2 and 3. Answer B (67\frac{6}{7}) might result from misapplying the relationship between consecutive binomial coefficients. Answer C (218\frac{21}{8}) is the reciprocal of the correct answer, suggesting a mix-up in which coefficient goes in the numerator versus denominator. Study tip: Always include all parts of each term when finding binomial coefficients—the binomial coefficient, the powers of both terms, and track your exponents carefully to match the desired power of the variable.

Question 2

The fourth term in the expansion of (ax+1x)n(ax + \frac{1}{x})^n is 160. Given that this term is independent of xx, find the value of aa.

  1. 2 (correct answer)
  2. 4
  3. 6
  4. 8
Explanation: This is a two-step problem. First, find nn, then find aa. The fourth term in the expansion corresponds to r=3r=3 in the formula Tr+1T_{r+1}. The general term is Tr+1=(nr)(ax)n−r(1x)r=(nr)an−rxn−rx−r=(nr)an−rxn−2rT_{r+1} = \binom{n}{r}(ax)^{n-r}(\frac{1}{x})^r = \binom{n}{r} a^{n-r} x^{n-r} x^{-r} = \binom{n}{r} a^{n-r} x^{n-2r}. For the fourth term, r=3r=3, so the term is T4=(n3)an−3xn−6T_4 = \binom{n}{3} a^{n-3} x^{n-6}. We are given that this term is independent of xx, which means the power of xx must be zero. n−6=0  ⟹  n=6n-6 = 0 \implies n=6. Now we know n=6n=6. The fourth term is (63)a6−3x6−6=(63)a3\binom{6}{3} a^{6-3} x^{6-6} = \binom{6}{3} a^3. We are given that this term is equal to 160. (63)=6×5×43×2×1=20\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20. So, 20a3=16020a^3 = 160. a3=16020=8a^3 = \frac{160}{20} = 8. a=83=2a = \sqrt[3]{8} = 2. Distractor analysis: D: 8 is the value of a3a^3, obtained by forgetting to take the cube root. C: 6 is the value of nn, not aa. B: 4 could be the result of a calculation error, such as miscalculating the binomial coefficient, e.g., if (63)\binom{6}{3} was thought to be 10, then 10a3=160  ⟹  a3=1610a^3 = 160 \implies a^3=16, leading to a=163a=\sqrt[3]{16} which is close to 2.5, but 4 is a more common guess.

Question 3

The sum of the coefficients in the expansion of (ax−b)5(ax-b)^5 is 32. The sum of the coefficients in the expansion of (bx+a)5(bx+a)^5 is 243. Given that aa and bb are real numbers, find the value of aa.

  1. 0.5
  2. 2
  3. 2.5 (correct answer)
  4. 3
Explanation: The sum of the coefficients of a polynomial is found by substituting 1 for all variables. For the first expansion, (ax−b)5(ax-b)^5, setting x=1x=1 gives the sum of coefficients as (a(1)−b)5=(a−b)5(a(1)-b)^5 = (a-b)^5. We are given this sum is 32. So, (a−b)5=32(a-b)^5 = 32. Taking the fifth root gives a−b=2a-b=2. For the second expansion, (bx+a)5(bx+a)^5, setting x=1x=1 gives the sum of coefficients as (b(1)+a)5=(b+a)5(b(1)+a)^5 = (b+a)^5. We are given this sum is 243. So, (b+a)5=243(b+a)^5 = 243. Taking the fifth root gives b+a=3b+a=3. We now have a system of two linear equations:
  1. a−b=2a-b=2
  2. a+b=3a+b=3 Adding the two equations gives 2a=52a = 5, so a=2.5a=2.5. Subtracting the first equation from the second gives 2b=12b = 1, so b=0.5b=0.5. The question asks for the value of aa.
Distractor analysis: A: 0.5 is the value of bb. B: 2 is the value of the expression a−ba-b. D: 3 is the value of the expression a+ba+b.

Question 4

What is the ratio of the coefficient of x7x^7 to the coefficient of x6x^6 in the expansion of (2x+3)10(2x+3)^{10}?

  1. 47\frac{4}{7}
  2. 67\frac{6}{7}
  3. 218\frac{21}{8}
  4. 821\frac{8}{21} (correct answer)
Explanation: When you encounter binomial expansion problems asking for ratios of coefficients, you're working with the binomial theorem: (a+b)n=∑k=0n(nk)an−kbk(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k For (2x+3)10(2x+3)^{10}, the general term is (10k)(2x)10−k(3)k=(10k)210−k3kx10−k\binom{10}{k}(2x)^{10-k}(3)^k = \binom{10}{k}2^{10-k}3^k x^{10-k} To find the coefficient of x7x^7, you need the term where 10−k=710-k = 7, so k=3k = 3: (103)2733=120⋅128⋅27=414,720\binom{10}{3}2^7 3^3 = 120 \cdot 128 \cdot 27 = 414,720 For the coefficient of x6x^6, you need 10−k=610-k = 6, so k=4k = 4: (104)2634=210⋅64⋅81=1,088,640\binom{10}{4}2^6 3^4 = 210 \cdot 64 \cdot 81 = 1,088,640 The ratio is: 414,7201,088,640=821\frac{414,720}{1,088,640} = \frac{8}{21} This confirms answer D is correct. Answer A (47\frac{4}{7}) likely comes from incorrectly using just the binomial coefficients (103)(104)=120210=47\frac{\binom{10}{3}}{\binom{10}{4}} = \frac{120}{210} = \frac{4}{7}, ignoring the powers of 2 and 3. Answer B (67\frac{6}{7}) might result from misapplying the relationship between consecutive binomial coefficients. Answer C (218\frac{21}{8}) is the reciprocal of the correct answer, suggesting a mix-up in which coefficient goes in the numerator versus denominator. Study tip: Always include all parts of each term when finding binomial coefficients—the binomial coefficient, the powers of both terms, and track your exponents carefully to match the desired power of the variable.

Question 5

The sum of the coefficients in the expansion of (ax−b)5(ax-b)^5 is 32. The sum of the coefficients in the expansion of (bx+a)5(bx+a)^5 is 243. Given that aa and bb are real numbers, find the value of aa.

  1. 0.5
  2. 2
  3. 2.5 (correct answer)
  4. 3
Explanation: The sum of the coefficients of a polynomial is found by substituting 1 for all variables. For the first expansion, (ax−b)5(ax-b)^5, setting x=1x=1 gives the sum of coefficients as (a(1)−b)5=(a−b)5(a(1)-b)^5 = (a-b)^5. We are given this sum is 32. So, (a−b)5=32(a-b)^5 = 32. Taking the fifth root gives a−b=2a-b=2. For the second expansion, (bx+a)5(bx+a)^5, setting x=1x=1 gives the sum of coefficients as (b(1)+a)5=(b+a)5(b(1)+a)^5 = (b+a)^5. We are given this sum is 243. So, (b+a)5=243(b+a)^5 = 243. Taking the fifth root gives b+a=3b+a=3. We now have a system of two linear equations:
  1. a−b=2a-b=2
  2. a+b=3a+b=3 Adding the two equations gives 2a=52a = 5, so a=2.5a=2.5. Subtracting the first equation from the second gives 2b=12b = 1, so b=0.5b=0.5. The question asks for the value of aa.
Distractor analysis: A: 0.5 is the value of bb. B: 2 is the value of the expression a−ba-b. D: 3 is the value of the expression a+ba+b.

Question 6

In the expansion of (x2−2x)6(x^2 - \frac{2}{x})^6, what is the constant term?

  1. -160
  2. 160
  3. 240 (correct answer)
  4. 480
Explanation: The general term in the expansion of (a+b)n(a+b)^n is given by Tr+1=(nr)an−rbrT_{r+1} = \binom{n}{r} a^{n-r} b^r. For the given expression, a=x2a = x^2, b=−2xb = -\frac{2}{x}, and n=6n=6. The general term is Tr+1=(6r)(x2)6−r(−2x)r=(6r)x12−2r(−2)rx−r=(6r)(−2)rx12−3rT_{r+1} = \binom{6}{r} (x^2)^{6-r} (-\frac{2}{x})^r = \binom{6}{r} x^{12-2r} (-2)^r x^{-r} = \binom{6}{r} (-2)^r x^{12-3r}. The constant term is the term independent of xx, which means the exponent of xx must be 0. So, we set 12−3r=012-3r = 0, which gives 3r=123r=12, so r=4r=4. The constant term is found by substituting r=4r=4 into the coefficient part of the general term: Constant term = (64)(−2)4=6!4!2!×16=15×16=240\binom{6}{4} (-2)^4 = \frac{6!}{4!2!} \times 16 = 15 \times 16 = 240. Distractor analysis: A: -160 is obtained if one incorrectly sets 12−2r=r  ⟹  3r=12  ⟹  r=412-2r=r \implies 3r=12 \implies r=4, but then calculates for r=3r=3: (63)(−2)3=20(−8)=−160\binom{6}{3}(-2)^3 = 20(-8)=-160. B: 160 is obtained if one uses r=3r=3 and makes a sign error: (63)(2)3=20(8)=160\binom{6}{3}(2)^3 = 20(8)=160. D: 480 is obtained from a calculation error, such as miscalculating the binomial coefficient (64)=30\binom{6}{4} = 30 instead of 15, leading to 30×16=48030 \times 16 = 480.

Question 7

What is the number of distinct terms in the full expansion of (x+y+z)5(x+y+z)^5?

  1. 6
  2. 10
  3. 21 (correct answer)
  4. 125
Explanation: The terms in the expansion of (x+y+z)5(x+y+z)^5 are of the form C⋅xaybzcC \cdot x^a y^b z^c, where a,b,ca, b, c are non-negative integers such that a+b+c=5a+b+c=5. The number of distinct terms is the number of non-negative integer solutions to the equation a+b+c=5a+b+c=5. This is a classic stars and bars problem. We have n=5n=5 'stars' (the total power) to distribute among k=3k=3 'bins' (the variables x, y, z). The number of ways to do this is given by the formula (n+k−1k−1)\binom{n+k-1}{k-1}. Number of terms = (5+3−13−1)=(72)\binom{5+3-1}{3-1} = \binom{7}{2}. (72)=7×62=21\binom{7}{2} = \frac{7 \times 6}{2} = 21. Therefore, there are 21 distinct terms in the expansion. Distractor analysis: A: 6 is the number of terms for an exponent of 2, (2+3−13−1)=(42)=6\binom{2+3-1}{3-1}=\binom{4}{2}=6. It is also 5+15+1, which is the number of terms in a binomial expansion to the power of 5. B: 10 is the value of (52)\binom{5}{2} or (53)\binom{5}{3}, a common mistake when applying combination formulas incorrectly. D: 125 would be 535^3, which is not related to this problem.

Question 8

What is the number of distinct terms in the full expansion of (x+y+z)5(x+y+z)^5?

  1. 6
  2. 10
  3. 21 (correct answer)
  4. 125
Explanation: The terms in the expansion of (x+y+z)5(x+y+z)^5 are of the form C⋅xaybzcC \cdot x^a y^b z^c, where a,b,ca, b, c are non-negative integers such that a+b+c=5a+b+c=5. The number of distinct terms is the number of non-negative integer solutions to the equation a+b+c=5a+b+c=5. This is a classic stars and bars problem. We have n=5n=5 'stars' (the total power) to distribute among k=3k=3 'bins' (the variables x, y, z). The number of ways to do this is given by the formula (n+k−1k−1)\binom{n+k-1}{k-1}. Number of terms = (5+3−13−1)=(72)\binom{5+3-1}{3-1} = \binom{7}{2}. (72)=7×62=21\binom{7}{2} = \frac{7 \times 6}{2} = 21. Therefore, there are 21 distinct terms in the expansion. Distractor analysis: A: 6 is the number of terms for an exponent of 2, (2+3−13−1)=(42)=6\binom{2+3-1}{3-1}=\binom{4}{2}=6. It is also 5+15+1, which is the number of terms in a binomial expansion to the power of 5. B: 10 is the value of (52)\binom{5}{2} or (53)\binom{5}{3}, a common mistake when applying combination formulas incorrectly. D: 125 would be 535^3, which is not related to this problem.

Question 9

The coefficient of x2x^2 in the expansion of (1+kx)n(1−2x)(1+kx)^n(1-2x) is zero. Find kk in terms of nn. (Assume n≥2n \ge 2).

  1. k=2n−1k = \frac{2}{n-1}
  2. k=4nk = \frac{4}{n}
  3. k=4(n−1)k = 4(n-1)
  4. k=4n−1k = \frac{4}{n-1} (correct answer)
Explanation: When you encounter a polynomial expansion where you need to find a specific coefficient, think systematically about how terms combine. This question tests your understanding of binomial expansion and coefficient matching. To find the coefficient of x2x^2 in (1+kx)n(1−2x)(1+kx)^n(1-2x), you need to expand both factors and collect like terms. Using the binomial theorem, (1+kx)n=1+nkx+(n2)(kx)2+…(1+kx)^n = 1 + nkx + \binom{n}{2}(kx)^2 + \ldots The coefficient of x2x^2 comes from two sources when you multiply by (1−2x)(1-2x):
  • The x2x^2 term from the first expansion times the constant term 1: (n2)k2⋅1=n(n−1)2k2\binom{n}{2}k^2 \cdot 1 = \frac{n(n-1)}{2}k^2
  • The xx term from the first expansion times (−2x)(-2x): nkx⋅(−2x)=−2nkx2nkx \cdot (-2x) = -2nkx^2
Setting the total coefficient equal to zero: n(n−1)2k2−2nk=0\frac{n(n-1)}{2}k^2 - 2nk = 0 Factor out nknk: nk((n−1)k2−2)=0nk\left(\frac{(n-1)k}{2} - 2\right) = 0 Since we want a non-trivial solution, (n−1)k2−2=0\frac{(n-1)k}{2} - 2 = 0, giving us k=4n−1k = \frac{4}{n-1}, which is answer D. Answer A gives 2n−1\frac{2}{n-1}, which is half the correct value. Answer B gives 4n\frac{4}{n}, which ignores the (n−1)(n-1) factor from the binomial coefficient. Answer C gives 4(n−1)4(n-1), which has the relationship inverted. Remember: when finding coefficients in polynomial products, systematically identify all ways the desired power can be formed, then set up your equation carefully. Double-check your binomial coefficient calculations.

Question 10

In the expansion of (x3−ax)12(x^3 - \frac{a}{x})^{12}, the term containing x16x^{16} has a coefficient of 25344. Find the value of aa.

  1. -2 (correct answer)
  2. 1
  3. 2
  4. ±2\pm 2
Explanation: The general term of the expansion is Tr+1=(12r)(x3)12−r(−ax)rT_{r+1} = \binom{12}{r} (x^3)^{12-r} (-\frac{a}{x})^r. Simplifying the term: Tr+1=(12r)x36−3r(−a)rx−r=(12r)(−a)rx36−4rT_{r+1} = \binom{12}{r} x^{36-3r} (-a)^r x^{-r} = \binom{12}{r} (-a)^r x^{36-4r}. We are looking for the term containing x16x^{16}, so we set the power of xx to 16: 36−4r=16  ⟹  4r=20  ⟹  r=536-4r = 16 \implies 4r = 20 \implies r=5. The coefficient of this term is (125)(−a)5\binom{12}{5}(-a)^5. First, calculate the binomial coefficient: (125)=12⋅11⋅10⋅9⋅85⋅4⋅3⋅2⋅1=792\binom{12}{5} = \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 792. The coefficient is 792(−a)5=−792a5792(-a)^5 = -792a^5. We are given that this coefficient is 25344. −792a5=25344-792a^5 = 25344. a5=−25344792=−32a^5 = -\frac{25344}{792} = -32. Taking the fifth root of both sides gives a=−2a = -2. Distractor analysis: C: a=2a=2 is the result of a sign error, solving a5=32a^5=32. D: a=±2a=\pm 2 incorrectly assumes that an odd root can yield two real solutions, similar to an even root. B: a=1a=1 is a guess or the result of significant calculation error.

Question 11

The sum of the coefficients in the expansion of (3x−1)n(3x-1)^n is 128. What is the value of nn?

  1. 5
  2. 6
  3. 7 (correct answer)
  4. 8
Explanation: To find the sum of the coefficients of any polynomial expansion, we substitute x=1x=1 into the expression. Let P(x)=(3x−1)nP(x) = (3x-1)^n. The sum of the coefficients is P(1)P(1). P(1)=(3(1)−1)n=(3−1)n=2nP(1) = (3(1)-1)^n = (3-1)^n = 2^n. We are given that the sum of the coefficients is 128. So, 2n=1282^n = 128. We know that 128=27128 = 2^7. Therefore, n=7n=7. Distractor analysis: A: 5 corresponds to 25=322^5=32. B: 6 corresponds to 26=642^6=64. D: 8 corresponds to 28=2562^8=256. These distractors test for errors in recalling powers of 2.

Question 12

In the expansion of (x2−2x)6(x^2 - \frac{2}{x})^6, what is the constant term?

  1. -160
  2. 160
  3. 240 (correct answer)
  4. 480
Explanation: The general term in the expansion of (a+b)n(a+b)^n is given by Tr+1=(nr)an−rbrT_{r+1} = \binom{n}{r} a^{n-r} b^r. For the given expression, a=x2a = x^2, b=−2xb = -\frac{2}{x}, and n=6n=6. The general term is Tr+1=(6r)(x2)6−r(−2x)r=(6r)x12−2r(−2)rx−r=(6r)(−2)rx12−3rT_{r+1} = \binom{6}{r} (x^2)^{6-r} (-\frac{2}{x})^r = \binom{6}{r} x^{12-2r} (-2)^r x^{-r} = \binom{6}{r} (-2)^r x^{12-3r}. The constant term is the term independent of xx, which means the exponent of xx must be 0. So, we set 12−3r=012-3r = 0, which gives 3r=123r=12, so r=4r=4. The constant term is found by substituting r=4r=4 into the coefficient part of the general term: Constant term = (64)(−2)4=6!4!2!×16=15×16=240\binom{6}{4} (-2)^4 = \frac{6!}{4!2!} \times 16 = 15 \times 16 = 240. Distractor analysis: A: -160 is obtained if one incorrectly sets 12−2r=r  ⟹  3r=12  ⟹  r=412-2r=r \implies 3r=12 \implies r=4, but then calculates for r=3r=3: (63)(−2)3=20(−8)=−160\binom{6}{3}(-2)^3 = 20(-8)=-160. B: 160 is obtained if one uses r=3r=3 and makes a sign error: (63)(2)3=20(8)=160\binom{6}{3}(2)^3 = 20(8)=160. D: 480 is obtained from a calculation error, such as miscalculating the binomial coefficient (64)=30\binom{6}{4} = 30 instead of 15, leading to 30×16=48030 \times 16 = 480.

Question 13

What is the coefficient of the x4x^4 term in the expansion of (1+x)6(1−x)(1+x)^6(1-x)?

  1. -20
  2. -5 (correct answer)
  3. 5
  4. 35
Explanation: The expansion of the product is (1−x)∑r=06(6r)xr(1-x) \sum_{r=0}^{6} \binom{6}{r}x^r. To find the coefficient of x4x^4, we consider the ways to obtain an x4x^4 term from the product:
  1. Multiply the constant term (1) from (1−x)(1-x) by the x4x^4 term from (1+x)6(1+x)^6.
  2. Multiply the xx term (−x-x) from (1−x)(1-x) by the x3x^3 term from (1+x)6(1+x)^6. Term 1: The x4x^4 term in (1+x)6(1+x)^6 has a coefficient of (64)\binom{6}{4}. So this part contributes 1×(64)=151 \times \binom{6}{4} = 15. Term 2: The x3x^3 term in (1+x)6(1+x)^6 has a coefficient of (63)\binom{6}{3}. So this part contributes −1×(63)=−20-1 \times \binom{6}{3} = -20. The total coefficient of x4x^4 is the sum of these contributions: 15−20=−515 - 20 = -5.
Distractor analysis: D: 35 is the result of adding the magnitudes of the coefficients instead of subtracting: 15+20=3515+20=35. C: 5 is the result of a sign error: −15+20=5-15+20=5. A: -20 is the contribution from the second part only.

Question 14

Find the term independent of xx in the expansion of (1−1x2)(x+2x)6(1 - \frac{1}{x^2})(x + \frac{2}{x})^6.

  1. 60
  2. 100 (correct answer)
  3. 160
  4. 220
Explanation: The expression is (1−x−2)×(x+2x−1)6(1 - x^{-2}) \times (x + 2x^{-1})^6. The constant term of the product is found by summing the products of terms from each factor that result in x0x^0. Let's find the general term of (x+2x)6(x + \frac{2}{x})^6: Tr+1=(6r)x6−r(2x)r=(6r)2rx6−2rT_{r+1} = \binom{6}{r} x^{6-r} (\frac{2}{x})^r = \binom{6}{r} 2^r x^{6-2r}. The constant term of the entire expression is given by: 1×(constant term of (x+2x)6)−x−2×(x2 term of (x+2x)6)1 \times (\text{constant term of } (x + \frac{2}{x})^6) - x^{-2} \times (x^2 \text{ term of } (x + \frac{2}{x})^6)
  1. Find the constant term of (x+2x)6(x + \frac{2}{x})^6. We need x6−2r=x0x^{6-2r} = x^0, so 6−2r=0  ⟹  r=36-2r=0 \implies r=3. The term is (63)23=20×8=160\binom{6}{3} 2^3 = 20 \times 8 = 160.
  2. Find the x2x^2 term of (x+2x)6(x + \frac{2}{x})^6. We need x6−2r=x2x^{6-2r} = x^2, so 6−2r=2  ⟹  2r=4  ⟹  r=26-2r=2 \implies 2r=4 \implies r=2. The term is (62)22x2=15×4x2=60x2\binom{6}{2} 2^2 x^2 = 15 \times 4 x^2 = 60x^2. The coefficient is 60.
  3. Combine these results. The constant term of the full expression is 1×160−1×60=1001 \times 160 - 1 \times 60 = 100.
Distractor analysis: C: 160 is the constant term of (x+2x)6(x + \frac{2}{x})^6 only, ignoring the (1−1x2)(1 - \frac{1}{x^2}) factor. A: 60 is the coefficient of the x2x^2 term in (x+2x)6(x + \frac{2}{x})^6. D: 220 is the result of adding the two components instead of subtracting: 160+60=220160 + 60 = 220.

Question 15

What is the power of xx in the term with the largest coefficient in the expansion of (1+3x)8(1+3x)^8?

  1. 4
  2. 5
  3. 6 (correct answer)
  4. 8
Explanation: Let CrC_r be the coefficient of the term xrx^r in the expansion of (1+3x)8(1+3x)^8. Then Cr=(8r)3rC_r = \binom{8}{r} 3^r. To find the largest coefficient, we find the value of rr for which Cr+1>CrC_{r+1} > C_r. Cr+1Cr>1  ⟹  (8r+1)3r+1(8r)3r>1\frac{C_{r+1}}{C_r} > 1 \implies \frac{\binom{8}{r+1} 3^{r+1}}{\binom{8}{r} 3^r} > 1 8!(r+1)!(7−r)!8!r!(8−r)!×3>1\frac{\frac{8!}{(r+1)!(7-r)!}}{\frac{8!}{r!(8-r)!}} \times 3 > 1 8−rr+1×3>1\frac{8-r}{r+1} \times 3 > 1 24−3r>r+124 - 3r > r+1 23>4r23 > 4r r<234=5.75r < \frac{23}{4} = 5.75. This inequality holds for r=0,1,2,3,4,5r = 0, 1, 2, 3, 4, 5. This means that the coefficients are increasing up to C6C_6 (since for r=5r=5, C6>C5C_6 > C_5). For r=6r=6, we have 6>5.756 > 5.75, so C7<C6C_7 < C_6. Thus, the largest coefficient occurs when r=6r=6. The power of xx is 6. Distractor analysis: A: 4 is the middle power, which would be correct for (1+x)8(1+x)^8 but not for (1+3x)8(1+3x)^8. B: 5 is the largest integer value of rr that satisfies r<5.75r < 5.75, a common misinterpretation of the result. D: 8 is the value of nn.

Question 16

The first three terms in the expansion of (1+ax)n(1+ax)^n in ascending powers of xx are 1−12x+60x21 - 12x + 60x^2. Find the values of aa and nn.

  1. n=6,a=2n=6, a=2
  2. n=6,a=−2n=6, a=-2 (correct answer)
  3. n=8,a=−1.5n=8, a=-1.5
  4. n=4,a=−3n=4, a=-3
Explanation: The expansion of (1+ax)n(1+ax)^n is given by (n0)+(n1)(ax)+(n2)(ax)2+...\binom{n}{0} + \binom{n}{1}(ax) + \binom{n}{2}(ax)^2 + ... This simplifies to 1+nax+n(n−1)2a2x2+...1 + nax + \frac{n(n-1)}{2}a^2x^2 + .... By comparing the coefficients with the given expansion 1−12x+60x21 - 12x + 60x^2, we can set up a system of equations:
  1. Coefficient of xx: na=−12na = -12
  2. Coefficient of x2x^2: n(n−1)2a2=60\frac{n(n-1)}{2}a^2 = 60 From equation (1), we have a=−12na = -\frac{12}{n}. Substitute this into equation (2): n(n−1)2(−12n)2=60\frac{n(n-1)}{2} (-\frac{12}{n})^2 = 60 n(n−1)2144n2=60\frac{n(n-1)}{2} \frac{144}{n^2} = 60 144(n−1)2n=60\frac{144(n-1)}{2n} = 60 72(n−1)=60n72(n-1) = 60n 72n−72=60n72n - 72 = 60n 12n=7212n = 72 n=6n = 6. Now substitute n=6n=6 back into equation (1) to find aa: 6a=−12  ⟹  a=−26a = -12 \implies a = -2. So, n=6n=6 and a=−2a=-2.
Distractor analysis: A: n=6,a=2n=6, a=2 results from a sign error, solving na=12na=12 instead of na=−12na=-12. C: For n=8,a=−1.5n=8, a=-1.5, na=8(−1.5)=−12na = 8(-1.5)=-12 is correct, but the x2x^2 coefficient is 8(7)2(−1.5)2=28(2.25)=63\frac{8(7)}{2}(-1.5)^2 = 28(2.25) = 63, which is close but not equal to 60. D: For n=4,a=−3n=4, a=-3, na=4(−3)=−12na = 4(-3)=-12 is correct, but the x2x^2 coefficient is 4(3)2(−3)2=6(9)=54\frac{4(3)}{2}(-3)^2 = 6(9) = 54, which is also close but not equal to 60.

Question 17

Find the coefficient of the x5x^5 term in the expansion of (2x−12)8(2x - \frac{1}{2})^8.

  1. -224 (correct answer)
  2. -14
  3. -7
  4. 224
Explanation: The general term is Tr+1=(8r)(2x)8−r(−12)rT_{r+1} = \binom{8}{r} (2x)^{8-r} (-\frac{1}{2})^r. To find the term with x5x^5, we set the power of xx to 5: 8−r=58-r = 5, which means r=3r=3. The term is (83)(2x)8−3(−12)3\binom{8}{3} (2x)^{8-3} (-\frac{1}{2})^3. The coefficient is (83)(2)5(−12)3\binom{8}{3} (2)^5 (-\frac{1}{2})^3. First, calculate the binomial coefficient: (83)=8×7×63×2×1=56\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56. Now, calculate the full coefficient: 56×25×(−18)=56×32×(−18)=56×(−4)=−22456 \times 2^5 \times (-\frac{1}{8}) = 56 \times 32 \times (-\frac{1}{8}) = 56 \times (-4) = -224. Distractor analysis: D: 224 is the result of ignoring the negative sign from (−12)3(-\frac{1}{2})^3. B: -14 is obtained by swapping the powers on the terms, i.e., calculating (83)(2)3(−12)5=56×8×(−132)=−14\binom{8}{3} (2)^3 (-\frac{1}{2})^5 = 56 \times 8 \times (-\frac{1}{32}) = -14. C: -7 is obtained by forgetting the coefficient 2 from the 2x2x term: (83)(−12)3=56×(−18)=−7\binom{8}{3} (-\frac{1}{2})^3 = 56 \times (-\frac{1}{8}) = -7.

Question 18

The sum of the coefficients in the expansion of (3x−1)n(3x-1)^n is 128. What is the value of nn?

  1. 5
  2. 6
  3. 7 (correct answer)
  4. 8
Explanation: To find the sum of the coefficients of any polynomial expansion, we substitute x=1x=1 into the expression. Let P(x)=(3x−1)nP(x) = (3x-1)^n. The sum of the coefficients is P(1)P(1). P(1)=(3(1)−1)n=(3−1)n=2nP(1) = (3(1)-1)^n = (3-1)^n = 2^n. We are given that the sum of the coefficients is 128. So, 2n=1282^n = 128. We know that 128=27128 = 2^7. Therefore, n=7n=7. Distractor analysis: A: 5 corresponds to 25=322^5=32. B: 6 corresponds to 26=642^6=64. D: 8 corresponds to 28=2562^8=256. These distractors test for errors in recalling powers of 2.

Question 19

In the expansion of (x+a)7(x+a)^7, the coefficient of the x4x^4 term is 2240. Find the value of aa, given a>0a>0.

  1. 2
  2. 4 (correct answer)
  3. 8
  4. 64
Explanation: The general term in the expansion of (x+a)7(x+a)^7 is Tr+1=(7r)x7−rarT_{r+1} = \binom{7}{r} x^{7-r} a^r. The term containing x4x^4 corresponds to 7−r=47-r=4, so r=3r=3. The term is (73)x4a3\binom{7}{3} x^4 a^3. The coefficient is (73)a3\binom{7}{3} a^3. First, calculate the binomial coefficient: (73)=7×6×53×2×1=35\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35. So the coefficient is 35a335a^3. We are given that this coefficient is 2240. 35a3=224035a^3 = 2240 a3=224035=448070=64a^3 = \frac{2240}{35} = \frac{4480}{70} = 64. a=643=4a = \sqrt[3]{64} = 4. Distractor analysis: A: 2 would result if a3=8a^3=8, which could come from a division error 2240/35=82240/35=8. C: 8 would result if the student thinks 64=8\sqrt{64}=8, confusing the cube root with the square root. D: 64 is the value of a3a^3, which results from forgetting to take the cube root.

Question 20

The first three terms in the expansion of (1+ax)n(1+ax)^n in ascending powers of xx are 1−12x+60x21 - 12x + 60x^2. Find the values of aa and nn.

  1. n=6,a=2n=6, a=2
  2. n=6,a=−2n=6, a=-2 (correct answer)
  3. n=8,a=−1.5n=8, a=-1.5
  4. n=4,a=−3n=4, a=-3
Explanation: The expansion of (1+ax)n(1+ax)^n is given by (n0)+(n1)(ax)+(n2)(ax)2+...\binom{n}{0} + \binom{n}{1}(ax) + \binom{n}{2}(ax)^2 + ... This simplifies to 1+nax+n(n−1)2a2x2+...1 + nax + \frac{n(n-1)}{2}a^2x^2 + .... By comparing the coefficients with the given expansion 1−12x+60x21 - 12x + 60x^2, we can set up a system of equations:
  1. Coefficient of xx: na=−12na = -12
  2. Coefficient of x2x^2: n(n−1)2a2=60\frac{n(n-1)}{2}a^2 = 60 From equation (1), we have a=−12na = -\frac{12}{n}. Substitute this into equation (2): n(n−1)2(−12n)2=60\frac{n(n-1)}{2} (-\frac{12}{n})^2 = 60 n(n−1)2144n2=60\frac{n(n-1)}{2} \frac{144}{n^2} = 60 144(n−1)2n=60\frac{144(n-1)}{2n} = 60 72(n−1)=60n72(n-1) = 60n 72n−72=60n72n - 72 = 60n 12n=7212n = 72 n=6n = 6. Now substitute n=6n=6 back into equation (1) to find aa: 6a=−12  ⟹  a=−26a = -12 \implies a = -2. So, n=6n=6 and a=−2a=-2.
Distractor analysis: A: n=6,a=2n=6, a=2 results from a sign error, solving na=12na=12 instead of na=−12na=-12. C: For n=8,a=−1.5n=8, a=-1.5, na=8(−1.5)=−12na = 8(-1.5)=-12 is correct, but the x2x^2 coefficient is 8(7)2(−1.5)2=28(2.25)=63\frac{8(7)}{2}(-1.5)^2 = 28(2.25) = 63, which is close but not equal to 60. D: For n=4,a=−3n=4, a=-3, na=4(−3)=−12na = 4(-3)=-12 is correct, but the x2x^2 coefficient is 4(3)2(−3)2=6(9)=54\frac{4(3)}{2}(-3)^2 = 6(9) = 54, which is also close but not equal to 60.