Problem 38 - Algebra 2

Question

Find the indicated term of each expansion. sixth term of \(\left(x-\frac{1}{2}\right)^{10}\)

Step-by-Step Solution

Answer

The sixth term is \(-\frac{63}{8} x^5\).

Step 1: Understand the Problem

We need to find the sixth term of the binomial expansion of \(\left(x-\frac{1}{2}\right)^{10}\). We will use the Binomial Theorem to help us.

Step 2: Write the Binomial Theorem

The Binomial Theorem states: \[(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]For our problem, \(a = x\), \(b = -\frac{1}{2}\), and \(n = 10\).

Step 3: Identify the Term Number

The sixth term of the expansion is given by \(T_{6} = \binom{10}{5} x^{10-5} \left(-\frac{1}{2}\right)^5\). This is because the k-th term in the expansion uses \(k-1\) as the index for \(b^k\), so we substitute \(k = 5\) for the sixth term.

Step 4: Calculate the Binomial Coefficient

Calculate the binomial coefficient \(\binom{10}{5}\):\[\binom{10}{5} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 252\]

Step 5: Calculate the Powers

Compute the powers:- \(x^{10-5} = x^5\)- \(\left(-\frac{1}{2}\right)^5 = -\left(\frac{1}{32}\right)\) (since \(-\frac{1}{2}\) raised to an odd power remains negative).

Step 6: Form the Expression for the Sixth Term

The sixth term \(T_{6}\) is:\[T_{6} = 252 \cdot x^5 \cdot \left(-\frac{1}{32}\right) = -\frac{252}{32} x^5\]This simplifies to \(-\frac{63}{8} x^5\).

Step 7: Simplify the Coefficient

Simplify \(-\frac{252}{32}\) by dividing both the numerator and the denominator by 4:\[-\frac{252}{32} = -\frac{63}{8}\]

Key Concepts

Binomial Expansion

The concept of binomial expansion is an essential tool in algebra. It deals with expanding expressions raised to a power, specifically of the form \((a + b)^n\). The Binomial Theorem helps us determine each term in this expansion by providing a clear formula. According to the theorem, an expression like \((a + b)^n\) can be expanded as a sum of terms involving combinations of the elements \(a\) and \(b\), each raised to various powers.
The general formula of the Binomial Theorem is given by:

• \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

This formula allows us to find any specific term in the expansion without needing to expand the entire expression. By substituting the values of \(a\), \(b\), and \(n\), we can solve for specific terms as needed in problems involving high powers. Understanding this concept makes it much easier to break down and solve complex expressions.

Binomial Coefficient

A binomial coefficient is a crucial part of the binomial expansion formula. It allows you to determine how many different ways you can combine a certain number of items from a larger set, represented as \(\binom{n}{k}\). This coefficient is often seen within the context of algebraic expansions involving powers.
Mathematically, the binomial coefficient \(\binom{n}{k}\) is calculated as:

• \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)

where \(n!\) (n factorial) is the product of all positive integers up to \(n\). These coefficients play a vital role in determining the weight of each term in a binomial expansion. For example, the problem above involves calculating \(\binom{10}{5}\) to find the sixth term in the expansion of \(\left(x - \frac{1}{2}\right)^{10}\). By understanding how to calculate these coefficients, you can effectively solve for terms in binomial expansions.

Powers in Binomial Expansion

Powers in a binomial expansion refer to the exponents applied to each component of the original binomial term \(a + b\). When expanding \((a + b)^n\), each term will involve different powers of both \(a\) and \(b\). The powers decrease for \(a\) from \(n\) to 0, while they increase for \(b\) from 0 to \(n\).
In the context of the given step-by-step solution, to find the sixth term of the binomial expansion \((x - \frac{1}{2})^{10}\), we identified the powers for \(x\) and \(-\frac{1}{2}\) as follows:

• The power of \(x\) is given by \(n-k\), where \(n\) is the total number of terms from the expansion (10 in this case) and \(k\) is the term number minus one (5 for the sixth term).
• Therefore, the power of \(x\) is \(10-5 = 5\).
• Similarly, the power of \(-\frac{1}{2}\) is \(k\), which is 5, leading to a component \((-\frac{1}{2})^5\).

The calculation of these powers is crucial in accurately determining each term's contribution to the overall expansion, ensuring a precise solution to the problem.