Problem 16 - Algebra 2 and Trigonometry

Question

In \(12-17,\) write the \(n\) th term of each binomial expansion. $$ (3 a+2 b)^{7}, n=6 $$

Step-by-Step Solution

Answer

The 6th term is \(6048a^2b^5\).

Step 1: Understand the Binomial Theorem

The binomial theorem states that \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\). Here, \((3a + 2b)^7\) is the expression given, with \(n=7\) representing the power of the expansion.

Step 2: Identify the Terms

For the binomial expression \((3a + 2b)^7\), \(x = 3a\) and \(y = 2b\). The general term \(T_{k+1}\) in the expansion is given by \(\binom{n}{k} (3a)^{n-k} (2b)^k\), where \(k\) determines the particular term.

Step 3: Determine the Position of the Term

The problem asks for the 6th term of the expansion. According to the binomial expansion formula, the \(n\)th term is actually the \(k+1\)th term in the formula. So for the 6th term, we set \(k=5\).

Step 4: Apply the Binomial Expansion Formula to Find the 6th Term

With \(k=5\), the 6th term \(T_6\) is calculated as: \[ T_6 = \binom{7}{5} (3a)^{7-5} (2b)^5 \]. Simplify each part.

Step 5: Calculate the Binomial Coefficient

Calculate \(\binom{7}{5}\) using the formula for binomial coefficients: \(\binom{7}{5} = \binom{7}{2} = \frac{7 \cdot 6}{2 \cdot 1} = 21\).

Step 6: Simplify the Powers

Simplify the powers: \( (3a)^{2} = 9a^2 \) and \( (2b)^5 = 32b^5 \).

Step 7: Multiply to Get Final Term

Now, combine these to find \(T_6\): \[ T_6 = 21 \cdot 9a^2 \cdot 32b^5 \]. Simplify it to get: \(T_6 = 6048a^2b^5\).

Key Concepts

Binomial Theorem

The Binomial Theorem is a key algebraic principle that allows us to expand expressions raised to a power into a sum of terms. When you have an expression in the form \((x + y)^n\), the theorem lets you expand this as a summation of terms. These terms involve both products and powers of the components \(x\) and \(y\). This expansion can be represented as: \[ (x+y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k \]In simpler terms, it tells you how to break down an expression that includes the sum of two items, both raised to a power, into a lengthy expression involving each part multiplied in different combinations. Each term in the expansion has its own binomial coefficient, which we will discuss further. Understanding the function of binomial theorem helps us simplify and calculate complex algebraic expressions with ease.
By recognizing and applying this theorem, we can systematically determine any term within the expansion instead of expanding everything manually.

Binomial Coefficients

Binomial coefficients play a crucial part in using the Binomial Theorem. They tell us the number of ways to select entities or terms. In the binomial expansion, each term is multiplied by a binomial coefficient, which is written as \(\binom{n}{k}\). This represents the number of combinations of \(k\) selections from \(n\) elements.

• Formula: \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\)
• This formula involves factorials, where a factorial (noted by an exclamation mark) means to multiply a series of descending natural numbers.
• In our example, to find \(\binom{7}{5}\), we calculated it as \(\binom{7}{2}\), giving us \(21\).

These coefficients help in scaling each term appropriately. Their understanding is also foundational for comprehending combinatory mathematics, which involves counting and arrangement principles. By mastering binomial coefficients, you enhance your skills in algebra significantly, as they are pivotal in determining the numeric multiplier for each term in the expansion.

Algebraic Expressions

Algebraic expressions are mathematical phrases combining numbers and variables using operations like addition, subtraction, multiplication, and division. They are used extensively in problems involving the Binomial Theorem. An expression such as \((3a + 2b)^7\) consists of two algebraic terms, \(3a\) and \(2b\), both of which can be anything from simple numbers to variable combinations. When working with the Binomial Theorem, our goal is to manipulate these algebraic components using algebraic operations:

• Identify the base terms in the expressions, such as \(x = 3a\) and \(y = 2b\) here.
• In the expansion, each term involves multiplying these base terms raised to different powers based on the position determined by \(k\).
• In step-by-step solutions, breaking down these expressions into manageable parts simplifies solving even quite complex algebraic problems.

Understanding how to handle such expressions effectively forms the backbone of successfully using and applying the Binomial Theorem in algebraic contexts. It allows for simplification and resolution of polynomial equations, enhancing problem-solving skills.