Determining The Coefficient Of X^30 In Polynomial Expansion

In the fascinating realm of algebra, polynomial expansions hold a significant position, especially when dealing with multinomial expressions. These expressions, characterized by multiple terms raised to various powers, often present intricate challenges in determining specific coefficients. Our focus here is to dissect the expansion of a particular multinomial expression and extract the coefficient of a specific term, x^30. This endeavor requires a meticulous approach, combining the principles of binomial theorem, multinomial theorem, and strategic algebraic manipulation. Understanding the nuances of these theorems is crucial for efficiently navigating such problems. The ability to identify and isolate the combinations of terms that contribute to the desired power of x is a testament to one's algebraic prowess. This article will serve as a comprehensive guide, walking you through each step of the process, elucidating the underlying concepts, and ultimately revealing the magnitude of the coefficient in question. Embark on this algebraic journey with us as we unravel the complexities and arrive at a definitive solution. So, let's delve into the heart of multinomial expansions and discover the elegant techniques that unlock their secrets. By the end of this exploration, you'll not only grasp the method for solving this particular problem but also gain a broader understanding of how to tackle similar algebraic challenges.

Problem Statement

Our mission is to find the coefficient of x^30 in the expansion of the expression:

(1 + 1/x)^6 (1 + x²⁾7 (1 - x³⁾8, where x ≠ 0

Let's denote this coefficient as α, and our ultimate goal is to determine the absolute value of α, represented as |α|.

This problem plunges us into the realm of polynomial expansions, where we must carefully consider how different terms interact and combine to produce the desired power of x. The expression itself is a product of three binomial expressions, each raised to a specific power. This complexity necessitates a strategic approach, leveraging the binomial theorem and carefully accounting for the contributions of each binomial factor. The presence of both positive and negative terms, as well as fractional exponents, adds layers of intricacy to the problem. However, with a systematic methodology, we can navigate these challenges and arrive at the correct answer. It's not just about crunching numbers; it's about understanding the underlying structure of the expansion and identifying the key terms that matter. This requires a keen eye for detail and a solid grasp of algebraic principles. So, prepare to put your algebraic skills to the test as we embark on this journey to find the elusive coefficient of x^30.

Breaking Down the Expression

The given expression is a product of three binomial expressions. To tackle this, we'll analyze each part separately:

1. (1 + 1/x)^6: This term can be rewritten using the binomial theorem. The general term in its expansion is given by:

   ⁶Cᵣ (1)⁶⁻ʳ (1/x)ʳ = ⁶Cᵣ x⁻ʳ

   where ⁶Cᵣ represents the binomial coefficient, often read as "6 choose r". This coefficient quantifies the number of ways to choose r items from a set of 6 items. The term x⁻ʳ indicates that this part of the expression will contribute negative powers of x to the overall expansion. Understanding this is crucial because we need to balance these negative powers with positive powers from other parts of the expression to obtain the x³⁰ term. The index 'r' ranges from 0 to 6, meaning we have seven distinct terms in this expansion, each with a different power of x. Carefully considering these powers and their corresponding coefficients is essential for finding the combinations that lead to x³⁰. The binomial theorem provides the framework, but strategic application is the key to success.

2. (1 + x²)^7: Similarly, the general term in the expansion of this part is:

   ⁷Cₛ (1)⁷⁻ₛ (x²)ˢ = ⁷Cₛ x²ˢ

   Here, ⁷Cₛ represents the binomial coefficient "7 choose s", and the power of x is 2s. This part of the expression will contribute even powers of x to the expansion, ranging from x⁰ to x¹⁴. The index 's' varies from 0 to 7, giving us eight terms in the expansion. Notice that all the powers of x are even, which is a significant observation. It implies that when combining terms from this expansion with those from other parts, we need to ensure that the overall power of x remains an integer. This constraint helps us narrow down the possible combinations that contribute to the x³⁰ term. The even powers also suggest a certain symmetry in the terms, which can be exploited for simplification. Understanding this characteristic is a valuable step in solving the problem efficiently.

3. (1 - x³)^8: The general term here is:

   ⁸Cₜ (1)⁸⁻ᵗ (-x³)ᵗ = ⁸Cₜ (-1)ᵗ x³ᵗ

   where ⁸Cₜ is the binomial coefficient "8 choose t". This term contributes powers of x that are multiples of 3, ranging from x⁰ to x²⁴. The presence of (-1)ᵗ introduces alternating signs depending on the value of t. This alternating sign is an important detail to keep track of, as it affects the overall sign of the coefficient we are trying to find. The index 't' ranges from 0 to 8, yielding nine terms in this expansion. The powers of x being multiples of 3 provides another constraint that helps us identify the relevant combinations. The alternating signs add a layer of complexity, but they also offer an opportunity for cancellation and simplification. By carefully analyzing the powers of x and the corresponding signs, we can effectively navigate this part of the expression and its contribution to the final coefficient.

Identifying Contributing Terms

To obtain the term x³⁰, we need to find combinations of r, s, and t such that:

-r + 2s + 3t = 30

with the constraints:

• 0 ≤ r ≤ 6
• 0 ≤ s ≤ 7
• 0 ≤ t ≤ 8

This equation is the crux of the problem. It encapsulates the relationship between the powers of x from each binomial expansion and dictates the combinations that will ultimately contribute to the x³⁰ term. Solving this Diophantine equation, subject to the given constraints, is a critical step. It requires a blend of algebraic manipulation and logical reasoning. We need to systematically explore the possible values of r, s, and t, keeping in mind the limitations imposed by the binomial coefficients and the powers to which the binomials are raised. The constraints are not just mathematical restrictions; they reflect the physical limitations of the expansions themselves. We can't choose more terms than are available in each expansion. This equation serves as a filter, allowing us to sift through the myriad of possibilities and pinpoint the combinations that truly matter. The challenge lies in finding an efficient method to solve it, avoiding exhaustive enumeration and leveraging mathematical insights to streamline the process.

Solving the Equation

Let's analyze the equation -r + 2s + 3t = 30 under the given constraints. We'll systematically explore possible solutions:

We can rearrange the equation to isolate r:

r = 2s + 3t - 30

Since 0 ≤ r ≤ 6, we have:

0 ≤ 2s + 3t - 30 ≤ 6

This inequality gives us a tighter range to work with. We can further refine it by adding 30 to all parts:

30 ≤ 2s + 3t ≤ 36

Now, we can systematically explore the possible values of t and s that satisfy this inequality, keeping in mind the constraints 0 ≤ s ≤ 7 and 0 ≤ t ≤ 8. We'll start by considering the extreme values of t and work our way inwards. This approach allows us to narrow down the possibilities efficiently. For each value of t, we'll determine the range of s values that satisfy the inequality. Then, we'll check if these values of s also satisfy the original constraints. This methodical approach ensures that we don't miss any potential solutions and that we only consider valid combinations of r, s, and t. It's a process of elimination and refinement, guided by the mathematical constraints and the desire to find all possible solutions.

• Case 1: t = 8

  If t = 8, then 2s + 3(8) ≤ 36, which simplifies to 2s ≤ 12, or s ≤ 6. Also, 2s + 3(8) ≥ 30, which simplifies to 2s ≥ 6, or s ≥ 3. So, for t = 8, the possible values of s are 3, 4, 5, and 6.

  For each of these values of s, we can calculate the corresponding value of r using the equation r = 2s + 3t - 30:

  • s = 3: r = 2(3) + 3(8) - 30 = 0
  • s = 4: r = 2(4) + 3(8) - 30 = 2
  • s = 5: r = 2(5) + 3(8) - 30 = 4
  • s = 6: r = 2(6) + 3(8) - 30 = 6

  These solutions represent valid combinations of r, s, and t that contribute to the x³⁰ term. Each combination corresponds to a specific term in the overall expansion, and we'll need to consider their coefficients when calculating the final result. The systematic exploration of possible values, starting with extreme cases, has allowed us to efficiently identify these solutions. This approach minimizes the risk of missing any combinations and ensures that we have a complete picture of the contributing terms. The next step is to calculate the coefficients for each of these combinations and sum them up, taking into account the alternating signs from the (1 - x³)^8 expansion.

• Case 2: t = 7

  If t = 7, then 2s + 3(7) ≤ 36, which simplifies to 2s ≤ 15, or s ≤ 7. Also, 2s + 3(7) ≥ 30, which simplifies to 2s ≥ 9, or s ≥ 4.5. Since s must be an integer, the possible values of s are 5, 6, and 7.

  For each of these values of s, we calculate the corresponding value of r:

  • s = 5: r = 2(5) + 3(7) - 30 = 1
  • s = 6: r = 2(6) + 3(7) - 30 = 3
  • s = 7: r = 2(7) + 3(7) - 30 = 5

  These are additional valid solutions that we need to consider. They expand our set of contributing terms and highlight the importance of a thorough search for all possible combinations. The fact that we found three solutions for t = 7 underscores the complexity of the problem and the need for a systematic approach. Each of these solutions will contribute to the final coefficient, and we'll need to carefully account for their individual contributions, including the signs and binomial coefficients. By continuing this methodical exploration, we ensure that we capture all the pieces of the puzzle and can accurately determine the coefficient of x³⁰.

• Case 3: t = 6

  If t = 6, then 2s + 3(6) ≤ 36, which simplifies to 2s ≤ 18, or s ≤ 9. Also, 2s + 3(6) ≥ 30, which simplifies to 2s ≥ 12, or s ≥ 6. So, the possible values of s are 6 and 7.

  For each of these values of s, we calculate the corresponding value of r:

  • s = 6: r = 2(6) + 3(6) - 30 = 0
  • s = 7: r = 2(7) + 3(6) - 30 = 4

  These two solutions add to our collection of contributing terms. We're gradually building a complete picture of all the combinations that produce x³⁰. The consistency of our approach, systematically exploring the possible values of t and s, ensures that we don't overlook any potential solutions. Each new solution reinforces the importance of careful calculation and attention to detail. As we move closer to the final answer, the accuracy of our intermediate steps becomes paramount. The coefficients associated with these solutions will be combined with the others we've found, and any error in this process could propagate to the final result.

• Case 4: t ≤ 5

  If t ≤ 5, then 3t ≤ 15. To satisfy 2s + 3t ≥ 30, we would need 2s ≥ 15, or s ≥ 7.5. Since the maximum value for s is 7, there are no solutions for t ≤ 5.

  This conclusion is significant because it allows us to stop our search. We've effectively proven that there are no more solutions to be found. By establishing this boundary, we avoid unnecessary calculations and can focus on combining the solutions we've already identified. This type of logical deduction is a powerful tool in problem-solving. It allows us to eliminate possibilities and streamline our approach. The realization that t ≤ 5 yields no solutions is a key step in simplifying the problem and bringing us closer to the final answer. With this knowledge, we can confidently move on to calculating the coefficients and summing them up.

Tabulating the Solutions

Let's summarize the solutions we've found in a table:

This table provides a clear and concise overview of all the combinations of r, s, and t that contribute to the x³⁰ term. It's a valuable tool for organization and helps us avoid double-counting or missing any solutions. The systematic approach we've taken has allowed us to construct this comprehensive table, which is essential for the next step: calculating the coefficients. Each row in the table represents a unique combination of terms from the three binomial expansions, and we need to determine the contribution of each combination to the final coefficient of x³⁰. The table serves as a roadmap, guiding us through the calculations and ensuring that we account for all the relevant terms.

Calculating the Coefficient

The coefficient α is given by the sum of the products of the binomial coefficients for each solution:

α = Σ [⁶Cᵣ * ⁷Cₛ * ⁸Cₜ * (-1)ᵗ]

where the sum is taken over all the solutions (r, s, t) listed in the table.

This formula encapsulates the essence of the problem. It tells us exactly how to combine the contributions of each solution to obtain the final coefficient. The binomial coefficients, ⁶Cᵣ, ⁷Cₛ, and ⁸Cₜ, quantify the number of ways each term can be formed in its respective binomial expansion. The (-1)ᵗ factor accounts for the alternating signs arising from the (1 - x³)^8 expansion. The summation symbol (Σ) indicates that we need to add up the contributions from all the solutions listed in our table. This formula is not just a mathematical expression; it's a precise recipe for calculating the coefficient. It highlights the importance of each component: the binomial coefficients, the alternating signs, and the systematic summation over all solutions. The accuracy of our final answer hinges on the correct application of this formula and the careful evaluation of each term in the sum.

Now, let's calculate the coefficient for each solution:

1. (0, 3, 8): ⁶C₀ * ⁷C₃ * ⁸C₈ * (-1)⁸ = 1 * 35 * 1 * 1 = 35
2. (2, 4, 8): ⁶C₂ * ⁷C₄ * ⁸C₈ * (-1)⁸ = 15 * 35 * 1 * 1 = 525
3. (4, 5, 8): ⁶C₄ * ⁷C₅ * ⁸C₈ * (-1)⁸ = 15 * 21 * 1 * 1 = 315
4. (6, 6, 8): ⁶C₆ * ⁷C₆ * ⁸C₈ * (-1)⁸ = 1 * 7 * 1 * 1 = 7
5. (1, 5, 7): ⁶C₁ * ⁷C₅ * ⁸C₇ * (-1)⁷ = 6 * 21 * 8 * (-1) = -1008
6. (3, 6, 7): ⁶C₃ * ⁷C₆ * ⁸C₇ * (-1)⁷ = 20 * 7 * 8 * (-1) = -1120
7. (5, 7, 7): ⁶C₅ * ⁷C₇ * ⁸C₇ * (-1)⁷ = 6 * 1 * 8 * (-1) = -48
8. (0, 6, 6): ⁶C₀ * ⁷C₆ * ⁸C₆ * (-1)⁶ = 1 * 7 * 28 * 1 = 196
9. (4, 7, 6): ⁶C₄ * ⁷C₇ * ⁸C₆ * (-1)⁶ = 15 * 1 * 28 * 1 = 420

These calculations represent the individual contributions of each solution to the overall coefficient. We've meticulously evaluated each product of binomial coefficients and accounted for the alternating signs. The numerical values vary significantly, highlighting the importance of each term in the sum. The positive and negative values indicate the constructive and destructive interference of the different terms in the expansion. Now, the final step is to sum these individual contributions to obtain the value of α. The accuracy of this summation is crucial, as it directly determines the final answer. The process requires careful attention to signs and arithmetic, ensuring that we arrive at the correct value for the coefficient of x³⁰.

Now, summing these values:

α = 35 + 525 + 315 + 7 - 1008 - 1120 - 48 + 196 + 420 = -628

This is the value of the coefficient α. It represents the net contribution of all the terms that combine to produce x³⁰ in the expansion. The negative sign indicates that the negative terms in the sum outweigh the positive terms. This result is a testament to the complexity of the multinomial expansion and the delicate balance between positive and negative contributions. We've successfully navigated the intricacies of the problem and arrived at a numerical value for the coefficient. The next step is to find the absolute value of this coefficient, which will give us the magnitude of the coefficient without regard to its sign.

Finding the Absolute Value

Finally, we need to find the absolute value of α:

|α| = |-628| = 628

This is the final answer to the problem. The absolute value of the coefficient of x³⁰ in the expansion is 628. This result represents the magnitude of the coefficient, regardless of its sign. It's a single, definitive number that encapsulates the outcome of our algebraic journey. We've successfully dissected the problem, applied the principles of binomial and multinomial expansions, solved a Diophantine equation, and meticulously calculated the coefficient. The absolute value provides a clean and unambiguous answer, summarizing the overall contribution of the x³⁰ term in the expansion.

We have successfully determined that the absolute value of the coefficient of x³⁰ in the expansion of (1 + 1/x)⁶ (1 + x²)⁷ (1 - x³)⁸ is 628. This problem showcases the power of the binomial theorem and the importance of systematic problem-solving in algebra. The journey involved breaking down a complex expression into manageable parts, identifying relevant terms, solving a Diophantine equation, and carefully calculating the contributions of each term. The final answer, 628, is a testament to the effectiveness of our approach and the elegance of algebraic principles. This exercise not only provides a solution to a specific problem but also enhances our understanding of multinomial expansions and our ability to tackle similar challenges in the future. The skills and techniques we've employed, such as systematic exploration, logical deduction, and careful calculation, are transferable to a wide range of mathematical problems. This experience underscores the value of a methodical approach and the satisfaction of arriving at a precise and meaningful solution.