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Magazine Chapter 6: Problem 13
Use the binomial theorem to find the coefficient of \(x^{a} y^{b}\) in the expansion of \(\left(2 x^{3}-4 y^{2}\right)^{7}\) , where a) \(a=9, b=8\) b) \(a=8, b=9\) c) \(a=0, b=14\) d \(a=12, b=6\) e) \(a=18, b=2\)
Short Answer
Expert verified
a) 71680, b) No such term, c) -16384, d) -35840, e) -1792
Step by step solution
01
- Understand the Binomial Theorem
The Binomial Theorem states that for any two terms x and y and any positive integer n, \[ (x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \] For this exercise, x and y will be replaced by terms involving powers of x and y.
02
- Identify the Terms in the Binomial Expansion
Given the expression \((2x^3 - 4y^2)^7\), the terms are \(2x^3\) and \(-4y^2\). Rewrite the general term in the binomial expansion: \[ T_k = \binom{7}{k} (2x^3)^{7-k} (-4y^2)^k \]
03
- Simplify the General Term
Simplify \((2x^3)^{7-k}\) and \((-4y^2)^k\): \[ (2x^3)^{7-k} = 2^{7-k} x^{3(7-k)} \] and \[ (-4y^2)^k = (-1)^k 4^k y^{2k} \]
04
- Combine the Terms
Combine the simplified terms to get the general term: \[ T_k = \binom{7}{k} 2^{7-k} (-1)^k 4^k x^{3(7-k)} y^{2k} \] Which simplifies to: \[ T_k = \binom{7}{k} 2^{7-k} (-1)^k 4^k x^{21-3k} y^{2k} \]
05
Step 5a - Solve for \(a = 9, b = 8\)
Set the exponents equal to the given values: \(21 - 3k = 9\) and \(2k = 8\). \ Taking the second equation: \[ 2k = 8 \] \[ k = 4 \] Substitute into the first equation to check consistency: \[ 21 - 3(4) = 9 \] which is true, so k = 4. Now find the coefficient: \[ T_4 = \binom{7}{4} 2^{7-4} (-1)^4 4^4 x^{9} y^{8} \] \[ T_4 = \binom{7}{4} 2^3 4^4 x^9 y^8 \] \[ T_4 = 35 \times 8 \times 256 x^9 y^8 \] \[ T_4 = 71680 x^9 y^8 \] The coefficient is 71680.
06
Step 5b - Solve for \(a = 8, b = 9\)
Set the exponents equal to the given values: \(21 - 3k = 8\) and \(2k = 9\). \ Taking the second equation: \[ 2k = 9 \] This is not possible as k must be an integer. Therefore, there is no such term in the expansion.
07
Step 5c - Solve for \(a = 0, b = 14\)
Set the exponents equal to the given values: \(21 - 3k = 0\) and \(2k = 14\). \ Taking the second equation: \[ 2k = 14 \] \[ k = 7 \] Substitute into the first equation to check consistency: \[ 21 - 3(7) = 0 \] which is true, so k = 7. Now find the coefficient: \[ T_7 = \binom{7}{7} 2^{7-7} (-1)^7 4^7 x^{0} y^{14} \] \[ T_7 = 1 \times 1 \times (-1)^7 \times 16384 x^0 y^{14} \] \[ T_7 = -16384 y^{14} \] The coefficient is -16384.
08
Step 5d - Solve for \(a = 12, b = 6\)
Set the exponents equal to the given values: \(21 - 3k = 12\) and \(2k = 6\). \ Taking the second equation: \[ 2k = 6 \] \[ k = 3 \] Substitute into the first equation to check consistency: \[ 21 - 3(3) = 12 \] which is true, so k = 3. Now find the coefficient: \[ T_3 = \binom{7}{3} 2^{7-3} (-1)^3 4^3 x^{12} y^{6} \] \[ T_3 = \binom{7}{3} 2^4 (-1)^3 64 x^{12} y^{6} \] \[ T_3 = 35 \times 16 \times (-1) \times 64 x^{12} y^{6} \] \[ T_3 = -35840 x^{12} y^{6} \] The coefficient is -35840.
09
Step 5e - Solve for \(a = 18, b = 2\)
Set the exponents equal to the given values: \(21 - 3k = 18\) and \(2k = 2\). \ Taking the second equation: \[ 2k = 2 \] \[ k = 1 \] Substitute into the first equation to check consistency: \[ 21 - 3(1) = 18 \] which is true, so k = 1. Now find the coefficient: \[ T_1 = \binom{7}{1} 2^{7-1} (-1) 4 x^{18} y^{2} \] \[ T_1 = 7 \times 2^6 (-1) \times 4 x^{18} y^{2} \] \[ T_1 = 7 \times 64 \times (-4) x^{18} y^{2} \] \[ T_1 = -1792 x^{18} y^{2} \] The coefficient is -1792.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
binomial expansion
The binomial expansion is a powerful method used to expand expressions that are raised to a power. According to the Binomial Theorem, for any two terms, which we can denote as x and y, and any positive integer n, the expansion of \((x + y)^n\) can be written as a sum of terms of the form \[ \binom{n}{k} x^{n-k} y^k \]. Here, \(\binom{n}{k}\) represents the binomial coefficient, which is a numerical value calculated from n and k.
coefficients
Coefficients are the numerical factors in terms of algebraic expressions. In the binomial expansion, coefficients determine the weight or importance of each term in the expansion. For instance, in \((2x^3 - 4y^2)^7\), each term in the expanded form is multiplied by coefficients that involve powers of 2, 4, and possibly negative signs. These coefficients are determined using the formula of the general term in binomial expansion, which involves products of binomial coefficients \(\binom{n}{k}\) and the specific terms being raised to their respective powers.
polynomial terms
Polynomial terms refer to the individual components of a polynomial expression. In the context of binomial expansion, each term in the polynomial is the product of three key elements: the binomial coefficient \(\binom{n}{k}\), the power of the first term x, and the power of the second term y. For example, given the expression \((2x^3 - 4y^2)^7\), a general polynomial term in this expansion can be written as \(\binom{7}{k}(2x^3)^{7-k}(-4y^2)^k\).
binomial coefficients
Binomial coefficients, denoted as \(\binom{n}{k}\), are crucial elements of the binomial theorem. They are also known as combination numbers and are calculated using the formula \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \], where \(n!\) denotes factorial of n. These coefficients give us the number of ways to choose k items from n items, and they play a significant role in determining the weight of each term in the expanded form of a binomial expression.
