Unveiling Coefficients: X³ And X⁻¹³ In Rational Function Expansion

Hey guys! Let's dive into a fun math problem. We're gonna break down how to find the coefficients of specific terms when we multiply out a bunch of rational functions. It's all about the binomial theorem and a bit of clever thinking. The goal is to find the sum of the coefficients of $x^3$ and $x^{-13}$ in the expansion of a given expression. Let's get started!

Understanding the Problem: A Deep Dive into the Expansion

So, the question asks us to look at the expansion of the expression (1 + x)(1 - x^2)ig(1 + rac{3}{x} + rac{3}{x^2} + rac{1}{x^3}ig)^5, where $x$ isn't zero. Basically, we want to multiply this whole thing out and then pinpoint the terms that have $x^3$ and $x^{-13}$. After we find those terms, we add their coefficients together, and boom, we have our answer! This problem blends algebra with the binomial theorem and the handling of exponents. First, we'll understand the different parts of the expression, starting with $(1 + x)$ and $(1 - x^2)$. These are simple polynomials that, when multiplied, will interact with the other part. The more complex part is ig(1 + rac{3}{x} + rac{3}{x^2} + rac{1}{x^3}ig)^5. This is where the binomial theorem shines. Note that the expression inside the parentheses can be written as ig(1 + rac{3}{x} + rac{3}{x^2} + rac{1}{x^3}ig) = (1 + rac{1}{x})^3, which helps simplify the rest of our work. This simplifies things quite a bit, as the expression inside the parentheses is actually a perfect cube.

The binomial theorem, in its simplest form, tells us how to expand expressions like $(a + b)^n$. In our case, it's useful for dealing with the fifth power of the given expression. We'll use it to break down that term and then figure out which parts of it, when combined with the other factors, produce $x^3$ and $x^{-13}$. Remember that the binomial theorem is just a formula for expanding powers of binomials. In our case, we are not dealing with just binomials, but it can be applied to understand the behavior of powers of polynomials like the one in the problem. Each term will be a combination of powers of $x$ and constants. Our job is to find the terms that result in the specific powers of x we're looking for.

Now, let's think about the entire expression. We have $(1 + x)(1 - x^2)$, which, when multiplied, gives us terms with powers of x from 0 to 2. And then, we have the much more involved ig(1 + rac{3}{x} + rac{3}{x^2} + rac{1}{x^3}ig)^5. This is where the binomial theorem becomes super important. Combining these two parts, we need to find out which combinations will provide us with the terms of $x^3$ and $x^{-13}$. It will be quite the challenge, but we are ready for it! The core strategy involves simplifying each part, expanding where needed, and then looking for the combinations that give us the target powers of $x$.

Breaking Down the Expression: Simplification and Binomial Power

Okay, let's get our hands dirty and simplify the expression. We have:

(1 + x)(1 - x^2)ig(1 + rac{3}{x} + rac{3}{x^2} + rac{1}{x^3}ig)^5

As we mentioned before, the most complex part is the fraction part, that simplifies as (1 + rac{1}{x})^3. So, rewrite the expression as follows:

(1 + x)(1 - x^2)ig(1 + rac{1}{x}ig)^{15}

Now, let's expand $(1 + x)(1 - x^2)$. This gives us: $1 - x^2 + x - x^3$.

So, now our original expression looks like this:

(1 - x^2 + x - x^3)ig(1 + rac{1}{x}ig)^{15}

Our job now is to find the terms which result in $x^3$ and $x^{-13}$ after the expression is multiplied. The key here is recognizing how the powers of $x$ will combine. We need to analyze how the terms from $(1 - x^2 + x - x^3)$ will interact with the expansion of ig(1 + rac{1}{x}ig)^{15} to produce the desired powers of $x$. Remember, each term in the expansion of ig(1 + rac{1}{x}ig)^{15} will be of the form C(15, k) * (rac{1}{x})^k, where $C(15, k)$ is a binomial coefficient. So, we will have to look for the right values of $k$ so that we can obtain $x^3$ and $x^{-13}$. Let's look at the terms that can produce $x^3$. For the constant 1 from the first part, we need $x^3$ from the second part. For $x$ from the first part, we need $x^2$ from the second part, and so on. We also have $-x^2$ and $-x^3$. In the expression ig(1 + rac{1}{x}ig)^{15}, the general term can be written as C(15, k) * (rac{1}{x})^k. Now, we seek to get $x^3$ by multiplying by a constant. Thus, $k = -3$, which cannot happen. If we multiply $x$ by a term, we need $x^2$. Thus, $k = -2$, which also cannot happen. If we multiply $x^2$ by a term, we need $x^5$. Thus $k = -5$, which also cannot happen. If we multiply $x^3$ by a term, we need $x^6$. Thus $k = -6$, which also cannot happen. So, we don't have any $x^3$ term.

Now let's look for the $x^{-13}$ term. We can multiply the constant term 1 with the $x^{-13}$ term from the second part. This can be done when $k = 13$. The $x^{-2}$ from the first part can be multiplied by $x^{-11}$ term. Thus $k = 11$. The $x$ term from the first part can be multiplied by the $x^{-14}$ term. Then $k = 14$. Finally, the $x^3$ from the first part can be multiplied by the $x^{-16}$ term. Then $k = 16$. Thus, we only need to find the coefficients for $k = 13$ and $k = 11$. The terms are C(15, 13) * (rac{1}{x})^{13} and -x^2 * C(15, 11) * (rac{1}{x})^{11}. The coefficients for the $x^{-13}$ is C(15, 13) = rac{15!}{13! * 2!} = 105. The coefficient for the second term is -C(15, 11) = -rac{15!}{11! * 4!} = -1365. Hence, the final answer is $105 - 1365 = -1260$. The coefficient of $x^3$ is zero and the coefficient for $x^{-13}$ is -1260. Then the sum is -1260.

Calculating Coefficients: The Binomial Theorem in Action

Let's use the binomial theorem to get the exact coefficients we need. For the expression ig(1 + rac{1}{x}ig)^{15}, the general term is:

C(15, k) * (rac{1}{x})^k

To get $x^3$, we need to multiply each term by the terms $1, x, -x^2, -x^3$. However, as we analyzed before, we cannot get the powers to obtain $x^3$. Now, for $x^{-13}$, we know that the terms that will contribute are $1$ and $-x^2$, so let's focus on those. The term that gives us $x^{-13}$ from the expansion of ig(1 + rac{1}{x}ig)^{15} is when $k = 13$: C(15, 13) * (rac{1}{x})^{13} = 105 * x^{-13}.

Then, from the second term, we have $-x^2$ from the first part and $x^{-11}$ from the second part. Thus, C(15, 11) * (rac{1}{x})^{11} * (-x^2) = -1365 * x^{-11} * x^2 = -1365x^{-13}. Now, all we have to do is to add the coefficients for $x^{-13}$. This will give us $105 - 1365 = -1260$. Hence, the sum of the coefficients of $x^3$ and $x^{-13}$ is $-1260$.

Conclusion: The Final Answer and Key Takeaways

Alright, guys, we've solved it! The sum of the coefficients of $x^3$ and $x^{-13}$ in the expansion is -1260. We went through some interesting steps. First, we simplified the expression as much as we could. Then, we focused on applying the binomial theorem to break down the complex part. We carefully examined the interactions between terms to find the ones with the desired powers of $x$. It's really cool to see how the binomial theorem lets us expand these complicated expressions and find specific coefficients. The main point to remember is to simplify as much as you can, and carefully keep track of powers of $x$. Also, be careful when you combine the different parts of the expression, and always double-check your calculations to make sure you're on the right track. I hope you enjoyed it! This is a classic example of how math skills like the binomial theorem can be used to solve complex problems.