Problem 45 - Algebra and Trigonometry

Question

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

Step-by-Step Solution

Answer

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

Step 1: 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).

Step 2: 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 \({{9}\choose{3}} x^{9-3}\left(-\frac{1}{2}\right)^3\).

Step 3: Calculate the Fourth Term

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

Key Concepts

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.