Binomial Theorem - IB Mathematics Analysis and Approaches

Find the coefficient of the $x^5$ term in the expansion of $(2x - \frac{1}{2})^8$ .
Answer: -224 The general term is $T_{r+1} = \binom{8}{r} (2x)^{8-r} (-\frac{1}{2})^r$ . To find the term with $x^5$ , we set the power of $x$ to 5: $8-r = 5$ , which means $r=3$ . The term is $\binom{8}{3} (2x)^{8-3} (-\frac{1}{2})^3$ . The coefficient is $\binom{8}{3} (2)^5 (-\frac{1}{2})^3$ . First, calculate the binomial coefficient: $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ . Now, calculate the full coefficient: $56 \times 2^5 \times (-\frac{1}{8}) = 56 \times 32 \times (-\frac{1}{8}) = 56 \times (-4) = -224$ . Distractor analysis: D: 224 is the result of ignoring the negative sign from $(-\frac{1}{2})^3$ . B: -14 is obtained by swapping the powers on the terms, i.e., calculating $\binom{8}{3} (2)^3 (-\frac{1}{2})^5 = 56 \times 8 \times (-\frac{1}{32}) = -14$ . C: -7 is obtained by forgetting the coefficient 2 from the $2x$ term: $\binom{8}{3} (-\frac{1}{2})^3 = 56 \times (-\frac{1}{8}) = -7$ .
What is the power of $x$ in the term with the largest coefficient in the expansion of $(1+3x)^8$ ?
Answer: 6 Let $C_r$ be the coefficient of the term $x^r$ in the expansion of $(1+3x)^8$ . Then $C_r = \binom{8}{r} 3^r$ . To find the largest coefficient, we find the value of $r$ for which $C_{r+1} > C_r$ . $\frac{C_{r+1}}{C_r} > 1 \implies \frac{\binom{8}{r+1} 3^{r+1}}{\binom{8}{r} 3^r} > 1$ $\frac{\frac{8!}{(r+1)!(7-r)!}}{\frac{8!}{r!(8-r)!}} \times 3 > 1$ $\frac{8-r}{r+1} \times 3 > 1$ $24 - 3r > r+1$ $23 > 4r$ $r < \frac{23}{4} = 5.75$ . This inequality holds for $r = 0, 1, 2, 3, 4, 5$ . This means that the coefficients are increasing up to $C_6$ (since for $r=5$ , $C_6 > C_5$ ). For $r=6$ , we have $6 > 5.75$ , so $C_7 < C_6$ . Thus, the largest coefficient occurs when $r=6$ . The power of $x$ is 6. Distractor analysis: A: 4 is the middle power, which would be correct for $(1+x)^8$ but not for $(1+3x)^8$ . B: 5 is the largest integer value of $r$ that satisfies $r < 5.75$ , a common misinterpretation of the result. D: 8 is the value of $n$ .
What is the coefficient of the $x^4$ term in the expansion of $(1+x)^6(1-x)$ ?
Answer: -5 The expansion of the product is $(1-x) \sum_{r=0}^{6} \binom{6}{r}x^r$ . To find the coefficient of $x^4$ , we consider the ways to obtain an $x^4$ term from the product: 1. Multiply the constant term (1) from $(1-x)$ by the $x^4$ term from $(1+x)^6$ . 2. Multiply the $x$ term ( $-x$ ) from $(1-x)$ by the $x^3$ term from $(1+x)^6$ . Term 1: The $x^4$ term in $(1+x)^6$ has a coefficient of $\binom{6}{4}$ . So this part contributes $1 \times \binom{6}{4} = 15$ . Term 2: The $x^3$ term in $(1+x)^6$ has a coefficient of $\binom{6}{3}$ . So this part contributes $-1 \times \binom{6}{3} = -20$ . The total coefficient of $x^4$ is the sum of these contributions: $15 - 20 = -5$ . Distractor analysis: D: 35 is the result of adding the magnitudes of the coefficients instead of subtracting: $15+20=35$ . C: 5 is the result of a sign error: $-15+20=5$ . A: -20 is the contribution from the second part only.
What is the ratio of the coefficient of $x^7$ to the coefficient of $x^6$ in the expansion of $(2x+3)^{10}$ ?
Answer: $\frac{8}{21}$ When you encounter binomial expansion problems asking for ratios of coefficients, you're working with the binomial theorem: $(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k$ For $(2x+3)^{10}$ , the general term is $\binom{10}{k}(2x)^{10-k}(3)^k = \binom{10}{k}2^{10-k}3^k x^{10-k}$ To find the coefficient of $x^7$ , you need the term where $10-k = 7$ , so $k = 3$ : $\binom{10}{3}2^7 3^3 = 120 \cdot 128 \cdot 27 = 414,720$ For the coefficient of $x^6$ , you need $10-k = 6$ , so $k = 4$ : $\binom{10}{4}2^6 3^4 = 210 \cdot 64 \cdot 81 = 1,088,640$ The ratio is: $\frac{414,720}{1,088,640} = \frac{8}{21}$ This confirms answer D is correct. Answer A ( $\frac{4}{7}$ ) likely comes from incorrectly using just the binomial coefficients $\frac{\binom{10}{3}}{\binom{10}{4}} = \frac{120}{210} = \frac{4}{7}$ , ignoring the powers of 2 and 3. Answer B ( $\frac{6}{7}$ ) might result from misapplying the relationship between consecutive binomial coefficients. Answer C ( $\frac{21}{8}$ ) 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.
In the binomial expansion of $(1+x)^n$ , the coefficient of $x^3$ is four times the coefficient of $x^2$ . Find the value of $n$ .
Answer: 14 The coefficient of $x^k$ in the expansion of $(1+x)^n$ is given by $\binom{n}{k}$ . The coefficient of $x^3$ is $\binom{n}{3}$ . The coefficient of $x^2$ is $\binom{n}{2}$ . According to the problem statement, $\binom{n}{3} = 4 \binom{n}{2}$ . Writing out the binomial coefficients: $\frac{n!}{3!(n-3)!} = 4 \times \frac{n!}{2!(n-2)!}$ $\frac{n(n-1)(n-2)}{3 \times 2 \times 1} = 4 \times \frac{n(n-1)}{2 \times 1}$ Assuming $n \ge 3$ , we can divide both sides by $n(n-1)$ : $\frac{n-2}{6} = \frac{4}{2}$ $\frac{n-2}{6} = 2$ $n-2 = 12$ $n=14$ . Distractor analysis: A: 8 results from a factorial simplification error: $\frac{n-2}{3} = 2 \implies n-2=6 \implies n=8$ . D: 19 results from confusing the terms, e.g., setting the coefficient of $x^4$ to be 4 times the coefficient of $x^3$ : $\binom{n}{4} = 4\binom{n}{3} \implies \frac{n-3}{4} = 4 \implies n-3=16 \implies n=19$ . B: 11 is a plausible distractor without a clear error path but close to the correct answer.
In the expansion of $(x+a)^7$ , the coefficient of the $x^4$ term is 2240. Find the value of $a$ , given $a>0$ .
Answer: 4 The general term in the expansion of $(x+a)^7$ is $T_{r+1} = \binom{7}{r} x^{7-r} a^r$ . The term containing $x^4$ corresponds to $7-r=4$ , so $r=3$ . The term is $\binom{7}{3} x^4 a^3$ . The coefficient is $\binom{7}{3} a^3$ . First, calculate the binomial coefficient: $\binom{7}{3} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$ . So the coefficient is $35a^3$ . We are given that this coefficient is 2240. $35a^3 = 2240$ $a^3 = \frac{2240}{35} = \frac{4480}{70} = 64$ . $a = \sqrt[3]{64} = 4$ . Distractor analysis: A: 2 would result if $a^3=8$ , which could come from a division error $2240/35=8$ . C: 8 would result if the student thinks $\sqrt{64}=8$ , confusing the cube root with the square root. D: 64 is the value of $a^3$ , which results from forgetting to take the cube root.
The coefficient of $x^2$ in the expansion of $(1+kx)^n(1-2x)$ is zero. Find $k$ in terms of $n$ . (Assume $n \ge 2$ ).
Answer: $k = \frac{4}{n-1}$ 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 $x^2$ in $(1+kx)^n(1-2x)$ , you need to expand both factors and collect like terms. Using the binomial theorem, $(1+kx)^n = 1 + nkx + \binom{n}{2}(kx)^2 + \ldots$ The coefficient of $x^2$ comes from two sources when you multiply by $(1-2x)$ : - The $x^2$ term from the first expansion times the constant term 1: $\binom{n}{2}k^2 \cdot 1 = \frac{n(n-1)}{2}k^2$ - The $x$ term from the first expansion times $(-2x)$ : $nkx \cdot (-2x) = -2nkx^2$ Setting the total coefficient equal to zero: $\frac{n(n-1)}{2}k^2 - 2nk = 0$ Factor out $nk$ : $nk\left(\frac{(n-1)k}{2} - 2\right) = 0$ Since we want a non-trivial solution, $\frac{(n-1)k}{2} - 2 = 0$ , giving us $k = \frac{4}{n-1}$ , which is answer D. Answer A gives $\frac{2}{n-1}$ , which is half the correct value. Answer B gives $\frac{4}{n}$ , which ignores the $(n-1)$ factor from the binomial coefficient. Answer C gives $4(n-1)$ , 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.
The coefficients of the terms $x^4$ , $x^5$ , and $x^6$ in the expansion of $(1+x)^n$ are in arithmetic progression. Find the sum of the possible values of $n$ .
Answer: 21 The coefficients are $\binom{n}{4}$ , $\binom{n}{5}$ , and $\binom{n}{6}$ . Since they are in an arithmetic progression, the middle term is the average of the other two. $2 \binom{n}{5} = \binom{n}{4} + \binom{n}{6}$ . Divide by $\binom{n}{5}$ : $2 = \frac{\binom{n}{4}}{\binom{n}{5}} + \frac{\binom{n}{6}}{\binom{n}{5}}$ . Using the identity $\frac{\binom{n}{k-1}}{\binom{n}{k}} = \frac{k}{n-k+1}$ , we have: $\frac{\binom{n}{4}}{\binom{n}{5}} = \frac{5}{n-5+1} = \frac{5}{n-4}$ and $\frac{\binom{n}{6}}{\binom{n}{5}} = \frac{n-5}{6}$ . Substituting these into the equation: $2 = \frac{5}{n-4} + \frac{n-5}{6}$ . Multiply by $6(n-4)$ to clear denominators: $12(n-4) = 30 + (n-4)(n-5)$ $12n - 48 = 30 + n^2 - 9n + 20$ $12n - 48 = n^2 - 9n + 50$ $0 = n^2 - 21n + 98$ . This is a quadratic equation for $n$ . We can solve it by factoring. We need two numbers that multiply to 98 and add to 21. These are 7 and 14. $(n-7)(n-14) = 0$ . The possible values for $n$ are $n=7$ and $n=14$ . The sum of the possible values is $7 + 14 = 21$ . Distractor analysis: A: 7 is one of the possible values. B: 14 is the other possible value. D: 23 is a plausible calculation error.
Find the middle term in the expansion of $(x^2 - \frac{1}{2x})^{10}$ .
Answer: $-\frac{63}{8}x^5$ For an expansion with power $n=10$ , there are $n+1=11$ terms. The middle term is the $\frac{11+1}{2} = 6^{\text{th}}$ term, which corresponds to $r=5$ in the formula $T_{r+1}$ . The general term is $T_{r+1} = \binom{10}{r} (x^2)^{10-r} (-\frac{1}{2x})^r$ . For the 6th term, we set $r=5$ : $T_6 = \binom{10}{5} (x^2)^{10-5} (-\frac{1}{2x})^5$ $T_6 = \binom{10}{5} (x^2)^5 (-\frac{1}{32x^5})$ $\binom{10}{5} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 252$ . $T_6 = 252 \cdot x^{10} \cdot (-\frac{1}{32x^5}) = -\frac{252}{32} x^{10-5} = -\frac{63}{8} x^5$ . Distractor analysis: C: $\frac{63}{8}x^5$ is the result of ignoring the negative sign from the $(-\frac{1}{2x})^5$ term. B: $\frac{105}{32}x^2$ is the 7th term (r=6): $\binom{10}{6}(x^2)^4(-\frac{1}{2x})^6 = 210 x^8 \frac{1}{64x^6} = \frac{105}{32}x^2$ . D: $-252x^5$ is obtained by forgetting the coefficient $\frac{1}{2}$ from the second term in the binomial.
In the expansion of $(1+x+ax^2)^6$ , the coefficient of $x^2$ is 27. Find the value of $a$ .
Answer: 2 We can treat $(1+(x+ax^2))$ as a binomial and expand it. $(1+(x+ax^2))^6 = \binom{6}{0} + \binom{6}{1}(x+ax^2) + \binom{6}{2}(x+ax^2)^2 + ...$ We only need terms up to $x^2$ to find the coefficient of $x^2$ . $= 1 + 6(x+ax^2) + 15(x^2 + 2ax^3 + a^2x^4) + ...$ Distribute and collect terms with $x^2$ : From $6(x+ax^2)$ , we get $6ax^2$ . From $15(x+ax^2)^2 = 15(x^2 + ...)$ , we get $15x^2$ . Higher terms in the initial expansion, like $\binom{6}{3}(x+ax^2)^3$ , will not produce an $x^2$ term. The total coefficient of $x^2$ is the sum of the coefficients from these parts: $6a + 15$ . We are given that this coefficient is 27. $6a + 15 = 27$ $6a = 12$ $a = 2$ . Distractor analysis: C: 4.5 is obtained by only considering the $6ax^2$ term and ignoring the $15x^2$ term: $6a=27 \implies a=4.5$ . A: 1.8 is obtained by incorrectly assuming the coefficient is $\binom{6}{2}a = 15a$ . Then $15a=27 \implies a=1.8$ . D: 7 could arise from a calculation error like $(27+15)/6 = 7$ .
Find the term independent of $x$ in the expansion of $(1 - \frac{1}{x^2})(x + \frac{2}{x})^6$ .
Answer: 100 The expression is $(1 - x^{-2}) \times (x + 2x^{-1})^6$ . The constant term of the product is found by summing the products of terms from each factor that result in $x^0$ . Let's find the general term of $(x + \frac{2}{x})^6$ : $T_{r+1} = \binom{6}{r} x^{6-r} (\frac{2}{x})^r = \binom{6}{r} 2^r x^{6-2r}$ . The constant term of the entire expression is given by: $1 \times (\text{constant term of } (x + \frac{2}{x})^6) - x^{-2} \times (x^2 \text{ term of } (x + \frac{2}{x})^6)$ 1. Find the constant term of $(x + \frac{2}{x})^6$ . We need $x^{6-2r} = x^0$ , so $6-2r=0 \implies r=3$ . The term is $\binom{6}{3} 2^3 = 20 \times 8 = 160$ . 2. Find the $x^2$ term of $(x + \frac{2}{x})^6$ . We need $x^{6-2r} = x^2$ , so $6-2r=2 \implies 2r=4 \implies r=2$ . The term is $\binom{6}{2} 2^2 x^2 = 15 \times 4 x^2 = 60x^2$ . The coefficient is 60. 3. Combine these results. The constant term of the full expression is $1 \times 160 - 1 \times 60 = 100$ . Distractor analysis: C: 160 is the constant term of $(x + \frac{2}{x})^6$ only, ignoring the $(1 - \frac{1}{x^2})$ factor. A: 60 is the coefficient of the $x^2$ term in $(x + \frac{2}{x})^6$ . D: 220 is the result of adding the two components instead of subtracting: $160 + 60 = 220$ .
The sum of the coefficients in the expansion of $(3x-1)^n$ is 128. What is the value of $n$ ?
Answer: 7 To find the sum of the coefficients of any polynomial expansion, we substitute $x=1$ into the expression. Let $P(x) = (3x-1)^n$ . The sum of the coefficients is $P(1)$ . $P(1) = (3(1)-1)^n = (3-1)^n = 2^n$ . We are given that the sum of the coefficients is 128. So, $2^n = 128$ . We know that $128 = 2^7$ . Therefore, $n=7$ . Distractor analysis: A: 5 corresponds to $2^5=32$ . B: 6 corresponds to $2^6=64$ . D: 8 corresponds to $2^8=256$ . These distractors test for errors in recalling powers of 2.
The first three terms in the expansion of $(1+ax)^n$ in ascending powers of $x$ are $1 - 12x + 60x^2$ . Find the values of $a$ and $n$ .
Answer: $n=6, a=-2$ The expansion of $(1+ax)^n$ is given by $\binom{n}{0} + \binom{n}{1}(ax) + \binom{n}{2}(ax)^2 + ...$ This simplifies to $1 + nax + \frac{n(n-1)}{2}a^2x^2 + ...$ . By comparing the coefficients with the given expansion $1 - 12x + 60x^2$ , we can set up a system of equations: 1) Coefficient of $x$ : $na = -12$ 2) Coefficient of $x^2$ : $\frac{n(n-1)}{2}a^2 = 60$ From equation (1), we have $a = -\frac{12}{n}$ . Substitute this into equation (2): $\frac{n(n-1)}{2} (-\frac{12}{n})^2 = 60$ $\frac{n(n-1)}{2} \frac{144}{n^2} = 60$ $\frac{144(n-1)}{2n} = 60$ $72(n-1) = 60n$ $72n - 72 = 60n$ $12n = 72$ $n = 6$ . Now substitute $n=6$ back into equation (1) to find $a$ : $6a = -12 \implies a = -2$ . So, $n=6$ and $a=-2$ . Distractor analysis: A: $n=6, a=2$ results from a sign error, solving $na=12$ instead of $na=-12$ . C: For $n=8, a=-1.5$ , $na = 8(-1.5)=-12$ is correct, but the $x^2$ coefficient is $\frac{8(7)}{2}(-1.5)^2 = 28(2.25) = 63$ , which is close but not equal to 60. D: For $n=4, a=-3$ , $na = 4(-3)=-12$ is correct, but the $x^2$ coefficient is $\frac{4(3)}{2}(-3)^2 = 6(9) = 54$ , which is also close but not equal to 60.
Given that the expansion of $(1 + \frac{x}{2})^n$ begins with $1 + 4x + ...$ , find the coefficient of $x^3$ .
Answer: 7 This is a two-step problem. First, we must find the value of $n$ . The binomial expansion of $(1 + \frac{x}{2})^n$ starts as: $\binom{n}{0} + \binom{n}{1}(\frac{x}{2})^1 + ... = 1 + n(\frac{x}{2}) + ... = 1 + \frac{n}{2}x + ...$ We are given that the expansion begins with $1+4x+...$ . By comparing the coefficients of $x$ , we have: $\frac{n}{2} = 4 \implies n=8$ . Now, we find the coefficient of $x^3$ in the expansion of $(1 + \frac{x}{2})^8$ . The term containing $x^3$ is $\binom{8}{3}(\frac{x}{2})^3$ . The coefficient is $\binom{8}{3}(\frac{1}{2})^3 = \binom{8}{3} \times \frac{1}{8}$ . Calculate the binomial coefficient: $\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ . The coefficient of $x^3$ is $56 \times \frac{1}{8} = 7$ . Distractor analysis: D: 56 is the value of $\binom{8}{3}$ , obtained by forgetting to cube the $\frac{1}{2}$ factor. A: 4 is the coefficient of $x$ given in the problem. C: 8 is the value of $n$ .
In the expansion of $(x^3 - \frac{a}{x})^{12}$ , the term containing $x^{16}$ has a coefficient of 25344. Find the value of $a$ .
Answer: -2 The general term of the expansion is $T_{r+1} = \binom{12}{r} (x^3)^{12-r} (-\frac{a}{x})^r$ . Simplifying the term: $T_{r+1} = \binom{12}{r} x^{36-3r} (-a)^r x^{-r} = \binom{12}{r} (-a)^r x^{36-4r}$ . We are looking for the term containing $x^{16}$ , so we set the power of $x$ to 16: $36-4r = 16 \implies 4r = 20 \implies r=5$ . The coefficient of this term is $\binom{12}{5}(-a)^5$ . First, calculate the binomial coefficient: $\binom{12}{5} = \frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 792$ . The coefficient is $792(-a)^5 = -792a^5$ . We are given that this coefficient is 25344. $-792a^5 = 25344$ . $a^5 = -\frac{25344}{792} = -32$ . Taking the fifth root of both sides gives $a = -2$ . Distractor analysis: C: $a=2$ is the result of a sign error, solving $a^5=32$ . D: $a=\pm 2$ incorrectly assumes that an odd root can yield two real solutions, similar to an even root. B: $a=1$ is a guess or the result of significant calculation error.
The sum of the coefficients in the expansion of $(ax-b)^5$ is 32. The sum of the coefficients in the expansion of $(bx+a)^5$ is 243. Given that $a$ and $b$ are real numbers, find the value of $a$ .
Answer: 2.5 The sum of the coefficients of a polynomial is found by substituting 1 for all variables. For the first expansion, $(ax-b)^5$ , setting $x=1$ gives the sum of coefficients as $(a(1)-b)^5 = (a-b)^5$ . We are given this sum is 32. So, $(a-b)^5 = 32$ . Taking the fifth root gives $a-b=2$ . For the second expansion, $(bx+a)^5$ , setting $x=1$ gives the sum of coefficients as $(b(1)+a)^5 = (b+a)^5$ . We are given this sum is 243. So, $(b+a)^5 = 243$ . Taking the fifth root gives $b+a=3$ . We now have a system of two linear equations: 1) $a-b=2$ 2) $a+b=3$ Adding the two equations gives $2a = 5$ , so $a=2.5$ . Subtracting the first equation from the second gives $2b = 1$ , so $b=0.5$ . The question asks for the value of $a$ . Distractor analysis: A: 0.5 is the value of $b$ . B: 2 is the value of the expression $a-b$ . D: 3 is the value of the expression $a+b$ .
What is the number of distinct terms in the full expansion of $(x+y+z)^5$ ?
Answer: 21 The terms in the expansion of $(x+y+z)^5$ are of the form $C \cdot x^a y^b z^c$ , where $a, b, c$ are non-negative integers such that $a+b+c=5$ . The number of distinct terms is the number of non-negative integer solutions to the equation $a+b+c=5$ . This is a classic stars and bars problem. We have $n=5$ 'stars' (the total power) to distribute among $k=3$ 'bins' (the variables x, y, z). The number of ways to do this is given by the formula $\binom{n+k-1}{k-1}$ . Number of terms = $\binom{5+3-1}{3-1} = \binom{7}{2}$ . $\binom{7}{2} = \frac{7 \times 6}{2} = 21$ . Therefore, there are 21 distinct terms in the expansion. Distractor analysis: A: 6 is the number of terms for an exponent of 2, $\binom{2+3-1}{3-1}=\binom{4}{2}=6$ . It is also $5+1$ , which is the number of terms in a binomial expansion to the power of 5. B: 10 is the value of $\binom{5}{2}$ or $\binom{5}{3}$ , a common mistake when applying combination formulas incorrectly. D: 125 would be $5^3$ , which is not related to this problem.
The fourth term in the expansion of $(ax + \frac{1}{x})^n$ is 160. Given that this term is independent of $x$ , find the value of $a$ .
Answer: 2 This is a two-step problem. First, find $n$ , then find $a$ . The fourth term in the expansion corresponds to $r=3$ in the formula $T_{r+1}$ . The general term is $T_{r+1} = \binom{n}{r}(ax)^{n-r}(\frac{1}{x})^r = \binom{n}{r} a^{n-r} x^{n-r} x^{-r} = \binom{n}{r} a^{n-r} x^{n-2r}$ . For the fourth term, $r=3$ , so the term is $T_4 = \binom{n}{3} a^{n-3} x^{n-6}$ . We are given that this term is independent of $x$ , which means the power of $x$ must be zero. $n-6 = 0 \implies n=6$ . Now we know $n=6$ . The fourth term is $\binom{6}{3} a^{6-3} x^{6-6} = \binom{6}{3} a^3$ . We are given that this term is equal to 160. $\binom{6}{3} = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ . So, $20a^3 = 160$ . $a^3 = \frac{160}{20} = 8$ . $a = \sqrt[3]{8} = 2$ . Distractor analysis: D: 8 is the value of $a^3$ , obtained by forgetting to take the cube root. C: 6 is the value of $n$ , not $a$ . B: 4 could be the result of a calculation error, such as miscalculating the binomial coefficient, e.g., if $\binom{6}{3}$ was thought to be 10, then $10a^3 = 160 \implies a^3=16$ , leading to $a=\sqrt[3]{16}$ which is close to 2.5, but 4 is a more common guess.
In the expansion of $(x^2 - \frac{2}{x})^6$ , what is the constant term?
Answer: 240 The general term in the expansion of $(a+b)^n$ is given by $T_{r+1} = \binom{n}{r} a^{n-r} b^r$ . For the given expression, $a = x^2$ , $b = -\frac{2}{x}$ , and $n=6$ . The general term is $T_{r+1} = \binom{6}{r} (x^2)^{6-r} (-\frac{2}{x})^r = \binom{6}{r} x^{12-2r} (-2)^r x^{-r} = \binom{6}{r} (-2)^r x^{12-3r}$ . The constant term is the term independent of $x$ , which means the exponent of $x$ must be 0. So, we set $12-3r = 0$ , which gives $3r=12$ , so $r=4$ . The constant term is found by substituting $r=4$ into the coefficient part of the general term: Constant term = $\binom{6}{4} (-2)^4 = \frac{6!}{4!2!} \times 16 = 15 \times 16 = 240$ . Distractor analysis: A: -160 is obtained if one incorrectly sets $12-2r=r \implies 3r=12 \implies r=4$ , but then calculates for $r=3$ : $\binom{6}{3}(-2)^3 = 20(-8)=-160$ . B: 160 is obtained if one uses $r=3$ and makes a sign error: $\binom{6}{3}(2)^3 = 20(8)=160$ . D: 480 is obtained from a calculation error, such as miscalculating the binomial coefficient $\binom{6}{4} = 30$ instead of 15, leading to $30 \times 16 = 480$ .

Start this quiz