Coefficient Of X³y In (2y + 4x³)^4 Binomial Expansion?

Hey math enthusiasts! Today, we're diving into the fascinating world of binomial expansions to figure out the coefficient of the x³y term in the expansion of (2y + 4x³)^4. Sounds like a mouthful, right? But don't worry, we'll break it down step by step. So, grab your calculators and let's get started!

Understanding the Binomial Theorem

Before we jump into the specifics, let's quickly refresh our understanding of the Binomial Theorem. This theorem provides a formula for expanding expressions of the form (a + b)^n, where n is a non-negative integer. The general formula looks like this:

(a + b)^n = Σ (n choose k) * a^(n-k) * b^k

Where:

• (n choose k) is the binomial coefficient, also written as nCk or C(n, k), and it represents the number of ways to choose k items from a set of n items.
• Σ represents the summation from k = 0 to n.
• a and b are the terms within the binomial.
• n is the power to which the binomial is raised.
• k is the index of summation, which ranges from 0 to n.

The binomial coefficient can be calculated using the following formula:

(n choose k) = n! / (k! * (n - k)!)

Where n! (n factorial) is the product of all positive integers up to n. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120.

The Binomial Theorem is a powerful tool for expanding binomials, and it's fundamental to many areas of mathematics, including algebra, calculus, and probability. It allows us to systematically determine the coefficients and terms in the expansion of a binomial raised to a power, without having to manually multiply the binomial by itself multiple times. This is particularly useful when dealing with large powers, where manual expansion would be tedious and time-consuming.

In essence, the Binomial Theorem provides a roadmap for navigating the expansion process, ensuring that we capture all the terms and their corresponding coefficients accurately. It's a cornerstone of algebraic manipulation and a key concept for anyone looking to deepen their understanding of mathematical expressions and their properties. The beauty of the Binomial Theorem lies in its ability to transform a seemingly complex expansion problem into a manageable series of calculations, making it an indispensable tool in the mathematician's toolkit.

Applying the Binomial Theorem to Our Problem

Okay, now that we've got the Binomial Theorem fresh in our minds, let's apply it to our specific problem: finding the coefficient of the x³y term in the expansion of (2y + 4x³)^4.

In this case:

• a = 2y
• b = 4x³
• n = 4

We're looking for the term that contains x³y. This means we need to find the value of k in the Binomial Theorem formula that will give us x³ and y when we plug in our values for a, b, and n. Remember, the general term in the binomial expansion is given by:

(4 choose k) * (2y)^(4-k) * (4x³)^k

Let's analyze the powers of x and y in this general term. We need the power of x to be 3 and the power of y to be 1. Looking at the term (4x³)^k, we can see that the power of x will be 3k. To get x³, we need 3k = 3, which means k = 1. Now, let's check the power of y. In the term (2y)^(4-k), the power of y will be 4 - k. If k = 1, then the power of y is 4 - 1 = 3. Oops! This doesn't give us the y term we're looking for.

Let's rethink this. We need the power of y to be 1. So, we need 4 - k = 1, which gives us k = 3. Now, let's check the power of x. If k = 3, the power of x will be 3k = 3 * 3 = 9. This gives us x⁹, which is not what we want either!

Wait a minute... it seems like we made a mistake in our interpretation of the problem. We are looking for the term x³y, not just x³ and y separately. So, let's go back to the general term and analyze it carefully. We need to find a k such that the term contains x³y. Let's plug in k = 1 into the general term and see what we get:

(4 choose 1) * (2y)^(4-1) * (4x³)^1 = 4 * (2y)³ * (4x³) = 4 * 8y³ * 4x³ = 128x³y³

This gives us x³y³, which is not the term we're looking for. It seems like there's no direct value of k that will give us the exact term x³y. This is because the power of x in the term (4x³)^k will always be a multiple of 3, and the power of y in the term (2y)^(4-k) will always be an integer. Therefore, we can't get a term with x³ and y in the expansion.

Finding the Correct Term and Coefficient

Alright, let's get our heads in the game and pinpoint the exact term we need. We're after the coefficient of the x³y term in the expansion of (2y + 4x³)^4. Remember, the binomial theorem is our trusty tool here. It tells us how to expand expressions like this, and it goes something like this:

(a + b)^n = Σ (n choose k) * a^(n-k) * b^k

Where, as we discussed:

• (n choose k) is the binomial coefficient.
• a and b are the terms inside the binomial.
• n is the exponent.

In our case, a is 2y, b is 4x³, and n is 4. So, we need to find the term in the expansion that looks like x³y. This means we need to figure out the right value of k in the binomial theorem formula.

The general term in our expansion will be:

(4 choose k) * (2y)^(4-k) * (4x³)^k

Let's break this down. The power of x in this term comes from (4x³)^k, which gives us x^(3k). The power of y comes from (2y)^(4-k), which gives us y^(4-k). We want the term with x³y, so we need:

• 3k = 3 (the power of x should be 3)
• 4 - k = 1 (the power of y should be 1)

From the first equation, we get k = 1. Plugging this into the second equation, we get 4 - 1 = 3, which is not equal to 1. So, there seems to be a conflict here. Let's try another approach. What if we look at the powers directly? We need x³, which comes from (4x³)^k. This means k must be 1. And we need y, which comes from (2y)^(4-k). If k = 1, then the power of y is 4 - 1 = 3. So, we get y³, not y.

It seems like there's no value of k that will give us exactly x³y. This is a bit puzzling! It might indicate that the term x³y doesn't actually appear in the expansion. Let's think about the possible terms we can get. The powers of x will always be multiples of 3 (0, 3, 6, etc.) because of the 4x³ term. The powers of y will be integers from 0 to 4. So, we can have terms like y^4, x³y³, x^6y², and x^9y. But no x³y!

Calculating the Coefficient

Okay, let's calculate the coefficient using the binomial theorem. We've established that:

• a = 2y
• b = 4x³
• n = 4

The general term in the binomial expansion is:

(n choose k) * a^(n-k) * b^k

Plugging in our values, we get:

(4 choose k) * (2y)^(4-k) * (4x³)^k

Now, we need to find the value of k that corresponds to the x³y term. As we discussed earlier, the power of x in the term (4x³)^k is 3k, and the power of y in the term (2y)^(4-k) is 4 - k. We want the term with x³y, so we need to solve the following equations:

• 3k = 3
• 4 - k = 1

From the first equation, we get k = 1. However, plugging k = 1 into the second equation gives us 4 - 1 = 3, which is not equal to 1. This tells us that there is no value of k that will simultaneously give us the powers of x and y that we need for the x³y term.

This is a crucial observation! It means that the term x³y does not actually appear in the expansion of (2y + 4x³)^4. So, what does this mean for the coefficient of the x³y term? Well, if a term doesn't exist in the expansion, its coefficient is simply 0.

The Final Answer

So, after carefully analyzing the binomial expansion and applying the Binomial Theorem, we've reached a conclusion. The term x³y does not exist in the expansion of (2y + 4x³)^4. Therefore, the coefficient of the x³y term is 0.

That's it, guys! We've successfully navigated this binomial expansion problem. Remember, sometimes the most important thing is not just applying formulas, but also understanding the underlying concepts and recognizing when a term simply doesn't exist. Keep practicing, and you'll become a master of binomial expansions in no time! Keep your passion for mathematics burning bright, and you'll conquer any challenge that comes your way. Until next time, happy calculating!