Hey guys! Today, we're diving deep into the fascinating world of binomial expansion. We'll tackle two intriguing problems: finding the coefficient of a specific term in an expansion and identifying the term containing a particular power of a variable. So, buckle up and let's get started!
Problem 1: Cracking the Coefficient of x³ in (x^(1/2) + 3)^12
Finding coefficients in binomial expansions might seem daunting at first, but with the right tools, it becomes a breeze. Our mission here is to pinpoint the coefficient of x³ within the expansion of (x^(1/2) + 3)^12. To conquer this, we'll employ the powerful Binomial Theorem. This theorem, a cornerstone of algebra, provides a formula for expanding expressions of the form (a + b)^n. The Binomial Theorem states that:
(a + b)^n = Σ [n! / (k! * (n - k)!)] * a^(n - k) * b^k, where k ranges from 0 to n.
This formula might look a bit intimidating, but let's break it down. The symbol Σ represents summation, meaning we'll be adding up a series of terms. The term inside the summation involves factorials (denoted by the exclamation mark), which represent the product of all positive integers up to a given number (e.g., 5! = 5 * 4 * 3 * 2 * 1). The fraction n! / (k! * (n - k)!) is known as the binomial coefficient, often written as C(n, k) or "n choose k". It represents the number of ways to choose k items from a set of n items.
Now, let's apply this to our problem. In our case, a = x^(1/2), b = 3, and n = 12. We're hunting for the term with x³, so we need to figure out the value of k that will give us this power of x. Looking at the general term in the binomial expansion, a^(n - k) becomes (x(1/2))(12 - k). We want this to simplify to x³. Therefore:
(x(1/2))(12 - k) = x³
Using the rules of exponents, we can rewrite the left side as x^((12 - k) / 2). Now we have:
x^((12 - k) / 2) = x³
For these two expressions to be equal, the exponents must be equal. So:
(12 - k) / 2 = 3
Multiplying both sides by 2 gives us:
12 - k = 6
Solving for k, we get:
k = 6
Great! We've found the value of k that corresponds to the x³ term. Now, we can plug this value back into the Binomial Theorem formula to find the coefficient of this term. The term with x³ is:
[12! / (6! * (12 - 6)!)] * (x(1/2))(12 - 6) * 3^6
Let's simplify this step by step. First, the binomial coefficient:
12! / (6! * 6!) = (12 * 11 * 10 * 9 * 8 * 7) / (6 * 5 * 4 * 3 * 2 * 1) = 924
Next, the power of x:
(x(1/2))(12 - 6) = (x(1/2))6 = x³
And finally, the power of 3:
3^6 = 729
Putting it all together, the term with x³ is:
924 * x³ * 729 = 673,596x³
Therefore, the coefficient of x³ in the expansion of (x^(1/2) + 3)^12 is 673,596. That's a pretty big number, but we cracked it using the Binomial Theorem!
Problem 2: Unearthing the y¹⁰ Term in (x - y²)⁸
Now, let's switch gears and tackle another binomial expansion challenge. This time, we're tasked with finding the term that contains y¹⁰ in the expansion of (x - y²)^8. The approach here is similar to the previous problem, but with a slight twist.
Again, we'll rely on the Binomial Theorem. In this case, a = x, b = -y² (notice the negative sign!), and n = 8. Our goal is to find the term with y¹⁰. Looking at the general term in the binomial expansion, b^k becomes (-y²)^k. We want this to simplify to a term containing y¹⁰. Therefore:
(-y²)^k = ...y¹⁰
Using the rules of exponents, we can rewrite the left side as (-1)^k * y^(2k). Now we have:
(-1)^k * y^(2k) = ...y¹⁰
For the powers of y to match, we need:
2k = 10
Dividing both sides by 2, we get:
k = 5
Excellent! We've found the value of k that gives us the y¹⁰ term. Now, we can plug this value back into the Binomial Theorem formula to find the entire term. The term with y¹⁰ is:
[8! / (5! * (8 - 5)!)] * x^(8 - 5) * (-y²)^5
Let's simplify this. First, the binomial coefficient:
8! / (5! * 3!) = (8 * 7 * 6) / (3 * 2 * 1) = 56
Next, the power of x:
x^(8 - 5) = x³
And finally, the power of -y²:
(-y²)^5 = (-1)^5 * (y²)^5 = -y¹⁰
Putting it all together, the term with y¹⁰ is:
56 * x³ * (-y¹⁰) = -56x³y¹⁰
Therefore, the term containing y¹⁰ in the expansion of (x - y²)^8 is -56x³y¹⁰. Notice the negative sign! It's crucial to keep track of signs when dealing with binomial expansions.
Key Takeaways and Conclusion
So, guys, we've successfully navigated the world of binomial expansion and conquered two challenging problems. We've learned how to find the coefficient of a specific term and how to identify the term containing a particular power of a variable. The key to success lies in understanding and applying the Binomial Theorem. Remember to carefully identify a, b, and n, and don't forget to account for negative signs!
Binomial expansion is a powerful tool with applications in various fields, including probability, statistics, and calculus. Mastering this concept opens doors to solving a wide range of mathematical problems. Keep practicing, and you'll become a binomial expansion pro in no time!
I hope this explanation was clear and helpful. If you have any questions, feel free to ask. Keep exploring the fascinating world of mathematics!