Solving Binomial Expansion Coefficients A0 - A1 + A2 - A3 + A4 - A5 + A6
Introduction
In this article, we delve into the fascinating world of binomial expansions and explore how to extract specific information from them. Our main focus will be on understanding the coefficients that arise when we expand a binomial expression raised to a power. Specifically, we'll tackle the problem presented: Given the expansion of (x - 2y)⁶ = A₀x⁶ - A₁x⁵y¹ + A₂x⁴y² - A₃x³y³ + A₄x²y⁴ - A₅xy⁵ + A₆y⁶, our mission is to determine the value of the alternating sum of coefficients: A₀ - A₁ + A₂ - A₃ + A₄ - A₅ + A₆. This exploration will not only enhance your understanding of binomial theorem applications but also sharpen your algebraic manipulation skills. Let's embark on this mathematical journey together!
Understanding Binomial Expansion
The binomial theorem is a powerful tool for expanding expressions of the form (a + b)ⁿ, where 'n' is a non-negative integer. The theorem provides a systematic way to determine the coefficients and terms in the resulting expansion. In general, the binomial theorem states:
(a + b)ⁿ = Σ (ⁿCᵣ) * a^(n-r) * b^r
where the summation symbol (Σ) indicates that we sum over all values of 'r' from 0 to 'n', and ⁿCᵣ represents the binomial coefficient, often read as "n choose r". This coefficient can be calculated using the formula:
ⁿCᵣ = n! / (r! * (n-r)!)
where '!' denotes the factorial function (e.g., 5! = 5 * 4 * 3 * 2 * 1). Understanding the binomial theorem is crucial for efficiently expanding expressions and extracting relevant information, such as specific coefficients or terms. The coefficients ⁿCᵣ, also known as binomial coefficients, exhibit a symmetrical pattern, which can be visualized using Pascal's Triangle. Each number in Pascal's Triangle is the sum of the two numbers directly above it, providing a convenient way to determine binomial coefficients for small values of 'n'. For larger values, the factorial formula is generally used. Let's explore how this applies to our specific problem.
Applying the Binomial Theorem to (x - 2y)⁶
Now, let's apply the binomial theorem to the expression (x - 2y)⁶. Here, we have a = x, b = -2y, and n = 6. Substituting these values into the binomial theorem formula, we get:
(x - 2y)⁶ = Σ (⁶Cᵣ) * x^(6-r) * (-2y)^r
Expanding this summation, we obtain:
(x - 2y)⁶ = ⁶C₀ * x⁶ * (-2y)⁰ + ⁶C₁ * x⁵ * (-2y)¹ + ⁶C₂ * x⁴ * (-2y)² + ⁶C₃ * x³ * (-2y)³ + ⁶C₄ * x² * (-2y)⁴ + ⁶C₅ * x¹ * (-2y)⁵ + ⁶C₆ * x⁰ * (-2y)⁶
Next, we need to calculate the binomial coefficients and simplify the terms. Recall that ⁶Cᵣ = 6! / (r! * (6-r)!). Let's compute each coefficient:
- ⁶C₀ = 1
- ⁶C₁ = 6
- ⁶C₂ = 15
- ⁶C₃ = 20
- ⁶C₄ = 15
- ⁶C₅ = 6
- ⁶C₆ = 1
Substituting these values back into the expansion, we get:
(x - 2y)⁶ = 1 * x⁶ * 1 + 6 * x⁵ * (-2y) + 15 * x⁴ * (4y²) + 20 * x³ * (-8y³) + 15 * x² * (16y⁴) + 6 * x * (-32y⁵) + 1 * 1 * (64y⁶)
Simplifying further, we arrive at:
(x - 2y)⁶ = x⁶ - 12x⁵y + 60x⁴y² - 160x³y³ + 240x²y⁴ - 192xy⁵ + 64y⁶
Comparing this expansion with the given expression, (x - 2y)⁶ = A₀x⁶ - A₁x⁵y¹ + A₂x⁴y² - A₃x³y³ + A₄x²y⁴ - A₅xy⁵ + A₆y⁶, we can identify the coefficients:
- A₀ = 1
- A₁ = 12
- A₂ = 60
- A₃ = 160
- A₄ = 240
- A₅ = 192
- A₆ = 64
These coefficients are the key to solving our problem. The careful application of the binomial theorem has allowed us to extract these coefficients accurately. Now, we proceed to the final step: calculating the alternating sum.
Calculating the Alternating Sum: A₀ - A₁ + A₂ - A₃ + A₄ - A₅ + A₆
Our primary goal is to find the value of the alternating sum: A₀ - A₁ + A₂ - A₃ + A₄ - A₅ + A₆. We have already determined the values of the coefficients A₀, A₁, A₂, A₃, A₄, A₅, and A₆ from the binomial expansion of (x - 2y)⁶. Now, we simply substitute these values into the expression and perform the arithmetic.
Substituting the values, we get:
1 - 12 + 60 - 160 + 240 - 192 + 64
Let's perform the calculations step by step:
- 1 - 12 = -11
- -11 + 60 = 49
- 49 - 160 = -111
- -111 + 240 = 129
- 129 - 192 = -63
- -63 + 64 = 1
Therefore, the alternating sum A₀ - A₁ + A₂ - A₃ + A₄ - A₅ + A₆ equals 1. This result demonstrates how manipulating the binomial expansion and correctly identifying the coefficients allows us to solve complex algebraic problems efficiently. The alternating sum provides a unique way to synthesize the information contained within the binomial coefficients.
An Alternative Method: Substitution
There is an elegant alternative method to find the alternating sum by using substitution. We start with the given binomial expansion:
(x - 2y)⁶ = A₀x⁶ - A₁x⁵y + A₂x⁴y² - A₃x³y³ + A₄x²y⁴ - A₅xy⁵ + A₆y⁶
We want to find A₀ - A₁ + A₂ - A₃ + A₄ - A₅ + A₆. Notice that this expression can be obtained by setting x = 1 and y = 1 in the expansion, with alternating signs. To achieve the alternating signs, we can substitute y = 1 into the original expression and then evaluate the result. However, to get the alternating signs, we need to consider the original binomial (x - 2y)⁶ and choose appropriate values for x and y such that the terms alternate in sign.
Let's set x = 1. Then the expansion becomes:
(1 - 2y)⁶ = A₀ - A₁y + A₂y² - A₃y³ + A₄y⁴ - A₅y⁵ + A₆y⁶
Now, we need to choose a value for y that will make the terms alternate in sign. Setting y = 1 gives us:
(1 - 2(1))⁶ = A₀ - A₁ + A₂ - A₃ + A₄ - A₅ + A₆
Simplifying the left side:
(1 - 2)⁶ = (-1)⁶ = 1
Thus, A₀ - A₁ + A₂ - A₃ + A₄ - A₅ + A₆ = 1. This substitution method provides a quick and efficient way to find the alternating sum without explicitly calculating each coefficient. It highlights the power of strategic substitutions in simplifying complex algebraic expressions.
Conclusion
In summary, we have successfully determined the value of A₀ - A₁ + A₂ - A₃ + A₄ - A₅ + A₆, which equals 1, by expanding the binomial expression (x - 2y)⁶ and identifying the coefficients. We also explored an alternative method using substitution, which provided a more direct route to the answer. The problem showcased the versatility of the binomial theorem and the importance of understanding how to manipulate algebraic expressions effectively. Mastering these techniques is crucial for tackling more advanced mathematical problems and enhancing problem-solving skills in general. By understanding the binomial theorem and its applications, we can efficiently solve problems involving polynomial expansions and coefficient analysis. The ability to extract specific information from these expansions, such as the alternating sum of coefficients, demonstrates a deep understanding of algebraic principles and their practical applications.
