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Find the term indicated in each expansion. $$ \left(x-\frac{1}{2}\right)^{9} ; \text { fourth term } $$

Short Answer

Expert verified
The fourth term in the expansion of \(\left(x-\frac{1}{2}\right)^9\) is \(-21x^6\).

Step by step solution

01

Identify the variables in the Binomial Theorem

Identify \(a\), \(b\), \(n\), and \(r\). Here, \(a=x\), \(b=-\frac{1}{2}\), \(n=9\), and \(r=3\) (since we're looking for the fourth term, and \(r\) is one less than the term number).
02

Use the Binomial Theorem

Using the Binomial Theorem, the expression for the fourth term is given by \({{n}\choose{r}} a^{n-r}b^r\), which is equivalent to \({{9}\choose{3}} x^{9-3}\left(-\frac{1}{2}\right)^3\).
03

Calculate the Fourth Term

Plug in values and simplify the expression. The binomial coefficient \({{9}\choose{3}} = 84\). Therefore, the fourth term is \(84x^6\left(-\frac{1}{2}\right)^3=-21x^6\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Binomial Expansion
Binomial expansion is a powerful algebraic tool that allows us to expand expressions like \((a + b)^n\) into a sum of terms. Each term in this expansion is of the form \({{n}\choose{r}}a^{n-r}b^r\), where:
  • \(n\) is the power to which the binomial is raised,
  • \(r\) is the term number minus one,
  • \(a\) and \(b\) are the components of the binomial.
The process involves applying the Binomial Theorem, which systematically forms each term through a combination of binomial coefficients and powers of the individual elements in the binomial.
For example, using this theorem to find the fourth term in \(\left(x-\frac{1}{2}\right)^{9}\) involves using previously determined parts like \(a = x\), \(b = -\frac{1}{2}\), and \(n = 9\). The term we're interested in corresponds to \(r = 3\), since it is \(n+1\) due to the term’s position in sequence.
Binomial Coefficient
The binomial coefficient is a crucial component of the binomial expansion process. It is represented by \({{n}\choose{r}}\) and reads as "n choose r". This coefficient represents the number of ways to choose \(r\) items from \(n\) items without regard to order, and it is calculated as:\[{{n}\choose{r}} = \frac{n!}{r!(n-r)!}\]In the given exercise, to find the fourth term of the expansion, we calculated the binomial coefficient for \({{9}\choose{3}}\). Here, the factorial notation \(n!\) signifies the product of all positive integers up to \(n\). For instance, \(3! = 3 \times 2 \times 1 = 6\). This results in \(\frac{9!}{3! \times 6!} = 84\), signifying 84 ways to arrange or select the terms required for our specific case in the polynomial expansion.
Polynomial Expansion
When expanding a polynomial expression like \(\left(x-\frac{1}{2}\right)^{9}\), each term contributes to the final polynomial expansion. Each term's construction is guided by the binomial theorem:
  • The base \(x\) in its powers contributes to the terms, reducing from \(9\) down to \(0\) as \(r\) increases.
  • The other term, here \(-\frac{1}{2}\), is similarly raised to increasing powers \(r\) from \(0\) to \(9\).
In our particular exercise, the focus was on the fourth term, calculated as \(84x^6\left(-\frac{1}{2}\right)^3\).
This becomes \(-21x^6\) after multiplying the coefficients and simplifying the powers. Polynomial expansions help in thoroughly expressing powers of binomials as a sum of many terms, making complex computations manageable and the understanding of polynomial behavior more intuitive.

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