Binomial Expansion Coefficients And Constant Terms
Hey there, math enthusiasts! Ever found yourself staring blankly at a binomial expansion, wondering how to extract a specific coefficient or, even more mysteriously, the term independent of the variable? Fear not, because we're about to embark on a journey into the fascinating world of binomial theorem applications. We'll break down complex expansions into manageable steps, arming you with the knowledge to confidently tackle these problems. Let's dive in and unravel the secrets hidden within these mathematical expressions!
a) Finding the Coefficient of 'a' in (a + 3)^5
So, finding the coefficient of a in the expansion of (a + 3)^5 might seem daunting at first, but trust me, it's totally doable! The binomial theorem is our best friend here. Remember, the binomial theorem provides a formula for expanding expressions of the form (x + y)^n. It states that:
(x + y)^n = Σ [nCk * x^(n-k) * y^k]
where Σ represents the summation from k = 0 to n, and nCk is the binomial coefficient, also known as "n choose k", calculated as n! / (k! * (n-k)!). In our case, we want to find the term where the power of a is 1. This means we're looking for the term where (n - k) = 1. Since n = 5 in our problem (a + 3)^5, we need to find the value of k that satisfies 5 - k = 1. Solving for k, we get k = 4. This tells us that the term containing a will be the one where k = 4 in the binomial expansion.
Now, let's plug in the values into the binomial theorem formula. We have n = 5, k = 4, x = a, and y = 3. The term we're interested in is:
5C4 * a^(5-4) * 3^4
Let's break this down step by step. First, we calculate the binomial coefficient 5C4, which is 5! / (4! * 1!). This simplifies to (5 * 4 * 3 * 2 * 1) / ((4 * 3 * 2 * 1) * 1) = 5. Next, we have a^(5-4) which is simply a^1 or just a. Then, we calculate 3^4, which is 3 * 3 * 3 * 3 = 81. Now, we put it all together:
5 * a * 81 = 405a
Therefore, the coefficient of a in the expansion of (a + 3)^5 is 405. See? Not so scary after all! By carefully applying the binomial theorem and identifying the correct term, we were able to pinpoint the coefficient we needed. This methodical approach is key to mastering binomial expansions. So, the next time you encounter a similar problem, remember to break it down into smaller steps, identify the relevant term using the binomial theorem, and you'll be well on your way to finding the solution. This method not only works for this specific problem but is a powerful tool for tackling a wide range of binomial expansion questions.
b) Cracking the Code Coefficient of 'a' in (a - 3b)^3
Okay, let's tackle another one! This time, we're aiming to find the coefficient of a in the expansion of (a - 3b)^3. The binomial theorem is our trusty companion once again. We need to identify the term in the expansion that contains a. Just like before, we'll use the formula:
(x + y)^n = Σ [nCk * x^(n-k) * y^k]
where Σ is the summation from k = 0 to n, and nCk is the binomial coefficient. In this case, n = 3, x = a, and y = -3b. We're looking for the term with a raised to the power of 1, so we need to find k such that (n - k) = 1. With n = 3, we have 3 - k = 1, which gives us k = 2. This means the term we're interested in is the one where k = 2.
Now, let's plug in the values into the binomial theorem formula. We have:
3C2 * a^(3-2) * (-3b)^2
Let's break it down. First, we calculate the binomial coefficient 3C2, which is 3! / (2! * 1!). This simplifies to (3 * 2 * 1) / ((2 * 1) * 1) = 3. Next, we have a^(3-2), which is simply a^1 or a. Then, we need to calculate (-3b)^2. Remember that squaring a negative number results in a positive number. So, (-3b)^2 = (-3)^2 * b^2 = 9b^2. Now, let's put it all together:
3 * a * 9b^2 = 27ab^2
The term containing a is 27ab^2. Therefore, the coefficient of a in the expansion of (a - 3b)^3 is 27b^2. Notice that the coefficient isn't just a number this time; it includes the term b^2. This is because we had a term with 'b' in the original binomial. These problems emphasize the importance of careful calculation and attention to detail, especially when dealing with negative signs and variables within the binomial. By understanding the binomial theorem and applying it systematically, you can successfully navigate these expansions and extract the specific coefficients you need.
c) Unraveling the Coefficient of 'a' in (a/2 - 2/3)^6
Alright, let's crank up the complexity a notch! We're now tasked with finding the coefficient of a in the expansion of (a/2 - 2/3)^6. This looks a bit more intricate, but the same principles of the binomial theorem apply. Our goal remains the same: identify the term where a is raised to the power of 1. Let's revisit the binomial theorem formula:
(x + y)^n = Σ [nCk * x^(n-k) * y^k]
Here, n = 6, x = a/2, and y = -2/3. We need to find k such that (n - k) = 1. With n = 6, we get 6 - k = 1, which means k = 5. So, we're looking at the term where k = 5 in the expansion. Plugging in the values into the formula, we have:
6C5 * (a/2)^(6-5) * (-2/3)^5
Let's break this down step by step. First, we calculate the binomial coefficient 6C5, which is 6! / (5! * 1!). This simplifies to (6 * 5 * 4 * 3 * 2 * 1) / ((5 * 4 * 3 * 2 * 1) * 1) = 6. Next, we have (a/2)^(6-5), which is (a/2)^1 or simply a/2. Then, we calculate (-2/3)^5. Raising a negative number to an odd power results in a negative number. So, (-2/3)^5 = - (2^5 / 3^5) = - (32 / 243). Now, let's put it all together:
6 * (a/2) * (-32/243) = (6 * a * -32) / (2 * 243) = -192a / 486
We can simplify the fraction -192/486 by dividing both numerator and denominator by their greatest common divisor, which is 6. This gives us -32/81. Therefore, the term containing a is -32a/81, and the coefficient of a in the expansion of (a/2 - 2/3)^6 is -32/81. This problem demonstrates how the binomial theorem can handle fractions and negative numbers within the binomial. It's crucial to pay close attention to the signs and to simplify the resulting expressions. The systematic approach of identifying the correct term and then carefully calculating each component is the key to success in these more complex expansions.
7. Finding the Term Independent of x
Now, let's shift gears slightly and tackle a different kind of problem: finding the term independent of x in a binomial expansion. What does this mean? Well, it means we're looking for the term in the expansion that doesn't have any x in it – the constant term. This type of problem often involves binomials with x in both terms, but with different powers, allowing for potential cancellation.
a) Discovering the Constant Term in (x^3 - 1/x2)5
So, let's dive into our first example: finding the term independent of x in the expansion of (x^3 - 1/x2)5. This means we need to find the term where the powers of x cancel out, leaving us with a constant. Once again, the binomial theorem is our guiding light:
(x + y)^n = Σ [nCk * x^(n-k) * y^k]
In this case, n = 5, our first term (which we'll treat as "x" in the formula) is x^3, and our second term (which we'll treat as "y" in the formula) is -1/x^2. To find the term independent of x, we need to find a value of k such that the powers of x in the term cancel each other out. Let's write out the general term in the expansion:
5Ck * (x3)(5-k) * (-1/x2)k
Now, let's focus on the powers of x. We have x^(3*(5-k)) from the first term and x^(-2*k) from the second term (remember, 1/x^2 is the same as x^-2). For the term to be independent of x, the exponents must add up to zero:
3(5 - k) - 2k = 0
Let's solve this equation for k. Expanding the equation, we get:
15 - 3k - 2k = 0
Combining like terms, we have:
15 - 5k = 0
Adding 5k to both sides and then dividing by 5, we get:
k = 3
This tells us that the term where k = 3 will be independent of x. Now, let's plug k = 3 back into the general term formula:
5C3 * (x3)(5-3) * (-1/x2)3
First, we calculate the binomial coefficient 5C3, which is 5! / (3! * 2!). This simplifies to (5 * 4 * 3 * 2 * 1) / ((3 * 2 * 1) * (2 * 1)) = 10. Next, we have (x3)(5-3), which is (x3)2 = x^6. Then, we calculate (-1/x2)3. Since we're raising a negative number to an odd power, the result will be negative. So, (-1/x2)3 = -1/x^6. Now, let's put it all together:
10 * x^6 * (-1/x^6) = -10
The x^6 terms cancel out, leaving us with -10. Therefore, the term independent of x in the expansion of (x^3 - 1/x2)5 is -10. This problem illustrates the power of the binomial theorem in identifying specific terms within an expansion. By focusing on the exponents and setting up an equation, we were able to pinpoint the value of k that yielded the constant term. This approach is applicable to a wide range of problems involving finding terms independent of a variable in binomial expansions.
In conclusion, mastering binomial expansions involves understanding and applying the binomial theorem, carefully identifying the relevant terms, and paying close attention to signs and exponents. Whether you're finding a specific coefficient or the term independent of x, a systematic approach will lead you to the solution. So, keep practicing, and you'll become a binomial expansion whiz in no time!
