Chapter 8: Problem 41
Find the term indicated in each expansion. \((x-1)^{9} ;\) fifth term
Short Answer
Expert verified
The fifth term in the expansion of \((x-1)^9\) is \(126x^5\).
Step by step solution
01
Calculating the combination
First calculate the combination \(_9C_4\), which gives us the coefficient of the term. This is done using the formula for combinations, which is \(^nC_r = \frac{n!}{r!(n-r)!}\) where n is the size of the set and r is the subset size. To find \(_9C_4\), it's \(\frac{9!}{4!(9-4)!}\) = 126.
02
Power of the first term
Now calculate the power of the first term a in the binomial. According to the term finding formula, 'a' will have a power of \(n - k\). Here, n is 9 and k is 4. So, the power of 'a' is \(9 - 4\) = 5.
03
Power of the second term
Next, find the power of the second term b in the binomial. According to the term finding formula, 'b' will have the power of k. Here, k is 4, so the power of 'b' is 4.
04
Assemble the term
The final step is to put these values together. Using the formula, the 5th term is: \(_9C_4 \cdot x^{9-4} \cdot (-1)^4\). Plugging in our found values, it becomes \(126 \cdot x^5 \cdot 1 = 126x^5\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Combination Formula
The combination formula is a crucial tool in combinatorics, used to determine the number of ways you can choose a subset of items from a larger set. It's denoted as \(^nC_r\) and is calculated using the formula:
Let's break it down with an example from our problem: we need \(_9C_4\).
- \(^nC_r = \frac{n!}{r!(n-r)!}\)
Let's break it down with an example from our problem: we need \(_9C_4\).
- Calculate \(9! = 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)
- Calculate \(4! = 4 \times 3 \times 2 \times 1\) and \(5! = 5 \times 4 \times 3 \times 2 \times 1\)
- \(\frac{9!}{4! \times 5!} = \frac{362880}{24 \times 120} = 126\)
Binomial Theorem
The Binomial Theorem provides a method to expand expressions that are raised to a power, in the form of
In our exercise, we have \((x - 1)^9\), and we need the fifth term. The relevant values are \(n = 9\) and \(k = 4\). So, the term is:
- \((a + b)^n\)
- \(_nC_k \cdot a^{n-k} \cdot b^k\)
In our exercise, we have \((x - 1)^9\), and we need the fifth term. The relevant values are \(n = 9\) and \(k = 4\). So, the term is:
- \(_9C_4 \cdot x^{9-4} \cdot (-1)^4 = 126 \cdot x^5 \cdot 1 = 126x^5\)
Polynomial Expression
A polynomial expression is a sum of terms consisting of variables raised to whole number exponents and multiplied by coefficients. These expressions can model a wide variety of real-world phenomena and are structured as \(ax^n + bx^{n-1} + \ldots + z\), where each term resembles a "monomial."
In the context of the Binomial Theorem, each binomial raised to a power generates a polynomial as its expansion. For example, \((x-1)^9\) becomes a polynomial with multiple terms, each formed by different coefficients and powers of \(x\). This is vital for simplifying expressions and solving equations.
Particularly, binomials like \((x-1)^9\) can be transformed into a polynomial composed of terms like \(ax^m\), showing how powerful and adaptable polynomials are in both mathematics and its applications. Understanding polynomials paves the way to grasp more complex algebraic concepts, significantly enhancing problem-solving skills in mathematics.
In the context of the Binomial Theorem, each binomial raised to a power generates a polynomial as its expansion. For example, \((x-1)^9\) becomes a polynomial with multiple terms, each formed by different coefficients and powers of \(x\). This is vital for simplifying expressions and solving equations.
Particularly, binomials like \((x-1)^9\) can be transformed into a polynomial composed of terms like \(ax^m\), showing how powerful and adaptable polynomials are in both mathematics and its applications. Understanding polynomials paves the way to grasp more complex algebraic concepts, significantly enhancing problem-solving skills in mathematics.