Problem 28 - College Algebra with Corequisite Support

Question

For the following exercises, use the Binomial Theorem to write the first three terms of each binomial. $$ (2 a+4 b)^{7} $$

Step-by-Step Solution

Answer

The first three terms are \(128a^7\), \(7168a^6b\), and \(215040a^5b^2\).

Step 1: Understanding the Binomial Theorem

The Binomial Theorem states that \((x + y)^n = \sum_{k=0}^{n} C(n, k) x^{n-k} y^k\), where \(C(n, k)\) is the combination of \(n\) items taken \(k\) at a time. This will allow us to expand the binomial expression and find specific terms.

Step 2: Determine the General Term Formula

The general term in the expansion is given by: \[ T_k = C(n, k) (2a)^{n-k} (4b)^k \]Here, \(n = 7\) as given in the problem.

Step 3: Calculate the First Term (k = 0)

For \(k = 0\), the term is:\[ T_0 = C(7, 0) (2a)^{7-0} (4b)^{0} = 1 \times (2a)^7 = 128a^7 \]

Step 4: Calculate the Second Term (k = 1)

For \(k = 1\), the term is:\[ T_1 = C(7, 1) (2a)^{7-1} (4b)^1 = 7 \times (2a)^6 \times 4b = 7168a^6b \]

Step 5: Calculate the Third Term (k = 2)

For \(k = 2\), the term is:\[ T_2 = C(7, 2) (2a)^{7-2} (4b)^2 = 21 \times (2a)^5 \times 16b^2 = 215040a^5b^2 \]

Key Concepts

Binomial Expansion

The binomial expansion is a way to expand expressions that are raised to a power. For example, given a binomial expression like \((x + y)^n\), the binomial theorem helps express it as a sum of terms involving the powers of \(x\) and \(y\). This expansion is particularly useful when \(n\) is a positive integer.

The basic formula for the binomial expansion is:

• \((x + y)^n = \sum_{k=0}^{n} C(n, k) x^{n-k} y^k\)

The sum represents adding together a sequence of terms. Each term involves a combination coefficient, denoted \(C(n, k)\), which depends on both \(n\) and \(k\).

As an example, consider the expression \((2a + 4b)^7\). Applying the binomial theorem means breaking it into a sequence of terms, each of which involves powers of \(2a\) and \(4b\). This is powerful because it turns a power expression into something much more granular.

Combination Formula

To find the individual terms in a binomial expansion, we rely on the combination formula. The combination formula, \(C(n, k)\), tells us how many ways we can choose \(k\) items from \(n\) items, disregarding the order. It's crucial for calculating the coefficients of each term in the expanded form.

Here's the formula for combinations:

• \(C(n, k) = \frac{n!}{k!(n-k)!}\)

Where \(!\) denotes factorial, which is the product of an integer and all the integers below it.

In the binomial example \((2a + 4b)^7\), we use this formula to find coefficients like \(C(7, 0)\), \(C(7, 1)\), and \(C(7, 2)\) to determine the first few terms in the expansion. This method systematically unravels the coefficients that form the foundation of each term's expression.

General Term of Binomial Expansion

The general term of a binomial expansion enables us to find any specific term in the expansion without having to fully expand the entire expression. This is especially useful when we're only interested in particular terms.

The general term is given by the formula:

• \(T_k = C(n, k) (x)^{n-k} (y)^k\)

Where \(T_k\) is the \(k\)-th term. This formula gives the flexibility to plug in any \(k\) to find that precise term in the expansion.

In our example with \((2a + 4b)^7\), we had three specific terms to compute: \(T_0\), \(T_1\), and \(T_2\). By substituting values into the general term formula, we easily calculate each term's contribution to the expanded form. For instance, when \(k = 0\), the formula simplifies to the calculation\( C(7, 0) (2a)^{7-0} (4b)^0\) which yields \(128a^7\). This makes it straightforward to assess and construct the expansion based on the relevant powers and coefficients.