Problem 44 - College Algebra

Question

In Exercises 39-48, find the term indicated in each expansion. $$\left(x^{3}+y^{2}\right)^{8} ; \text { sixth term }$$

Step-by-Step Solution

Answer

The sixth term in the expansion of \((x^{3}+y^{2})^{8}\) is \(56x^{9}y^{10}\)

Step 1: Identify the variables in our expression

In the given expression \((x^{3}+y^{2})^{8}\), we can identify: \(a = x^3\), \(b = y^2\), and \(n = 8\)

Step 2: Apply the Binomial theorem

The Binomial theorem tells us the coefficients of the terms when a binomial expression is expanded. The theorem has the formula: \((a+b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k}b^{k}\).

Step 3: Find the sixth term

The general term of binomial expansion is given as \(T_{k+1} = ^nC_k * a^{n-k} * b^{k}\). Here we are finding the sixth term, so \(k=6-1=5\). After substitution, we get the sixth term as \(T_{6} = ^8C_5 * (x^{3})^{8-5} * (y^{2})^{5}\)

Step 4: Simplify the Term

We calculate the Binomial Coefficient ^8C_5 = 56. Then we simplify the powers of \(x^3\) and \(y^2\) to get \(x^{9}\) and \(y^{10}\). Therefore the sixth term is \(56x^{9}y^{10}\)

Key Concepts

Binomial Expansion

The concept of binomial expansion is crucial when dealing with expressions raised to a power, like \((x^3 + y^2)^8\). This is governed by the Binomial Theorem, which provides a way to expand binomial expressions without manual multiplication.

The Binomial Theorem states:

• The expression \((a+b)^n\) can be expanded using the formula: \((a+b)^n = \sum_{k=0}^{n} {n \choose k} a^{n-k} b^k\).
• Each term in the expansion has a binomial coefficient \({n \choose k}\), calculated as \(\frac{n!}{k!(n-k)!}\).
• The expansion includes terms starting from \(a^n\) or \(b^n\), decreasing and increasing their powers by 1 respectively, as \(k\) increases.

This theorem simplifies calculating each term separately and is less error-prone than manual distribution.

Coefficient Calculation

The binomial coefficient, \({n \choose k}\), is an essential part of each term in a binomial expansion. It determines how many different ways we can choose \(k\) elements from a set of \(n\) elements and is central to calculating the coefficient of terms in the expansion.

In our problem \((x^3 + y^2)^8\), we need to calculate the coefficient for the sixth term. Here, based on the general form of the expansion, we use \(k=5\) (since the sixth term is \(T_6\)).

The coefficient is calculated as follows:

• Calculate \({8 \choose 5}\), the binomial coefficient for \(k=5\).
• This results in \({8 \choose 5} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56\).

This coefficient is then used to determine the sixth term in the expansion.

Power Simplification

Simplifying powers in binomial expansion is straightforward once you understand the basic principles. The power simplifies the elements within each term by applying the exponents separately to each component of the binomial terms.

In our example \((x^3 + y^2)^8\), the power of each variable is determined using the general form \(T_{k+1}\). For the sixth term:

• We calculate the powers of each variable: \((x^3)^{8-5} = (x^3)^3 = x^{9}\) and \((y^2)^5 = y^{10}\).
• Apply these exponents to simplify each binomial component within the term.

This type of simplification is crucial for computing the actual terms in the expression and requires careful attention to exponents.

General Term

The general term in a binomial expansion is derived using a formula that helps identify any term within the expansion sequence. The formula is:

• \(T_{k+1} = {n \choose k} a^{n-k} b^k\)
• This formula gives the \(k+1\)th term by using the binomial coefficient and the powers adjusted for the position \(k\).

For example, for the expression \((x^3 + y^2)^8\) and finding the sixth term:

• Substitute \(k = 5\), hence getting \(T_6\).
• \(T_6 = {8 \choose 5} (x^3)^{8-5} (y^2)^5\)

This approach is systematic and aids in pinpointing exact terms without expanding the whole sequence.