Problem 37 - Algebra and Trigonometry

Question

Use the Binomial Theorem to find the indicated coefficient or term. The 3 rd term in the expansion of \((3 x-2)^{9}\)

Step-by-Step Solution

Answer

The 3rd term is \(314928 x^7\).

Step 1: Identify the Formula for the Binomial Theorem

The Binomial Theorem states that: \[(a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\]Here, we need to identify the values for \(a\), \(b\), and \(n\). For \((3x - 2)^9\), we have \(a = 3x\), \(b = -2\), and \(n = 9\).

Step 2: Apply the Binomial Theorem to Find the Specified Term

The term in the expansion is given by: \[T_{k+1} = \binom{n}{k} a^{n-k} b^k\]To find the 3rd term, set \(k = 2\) (since term number is \(k+1\)): \[T_3 = \binom{9}{2} (3x)^{9-2} (-2)^2\]

Step 3: Calculate the Binomial Coefficient

Calculate \( \binom{9}{2} \): \[ \binom{9}{2} = \frac{9!}{2!(9-2)!} = \frac{9!}{2!7!} = \frac{9 \times 8}{2 \times 1} = 36\]

Step 4: Calculate the Powers

Calculate \((3x)^{7}\) and \((-2)^2\):\[ (3x)^7 = 3^7 x^7 = 2187 x^7 \]\((-2)^2 = 4\)

Step 5: Combine All Parts to Get the Term

Multiply all parts together to get the third term:\[T_3 = 36 \times 2187 x^7 \times 4\]Carry out the multiplication:\[T_3 = 36 \times 2187 \times 4 x^7 = 314928 x^7\]

Key Concepts

expansion

The Binomial Theorem allows us to expand expressions of the form \( (a+b)^n \) into a sum of terms. Each term in the expansion is derived using a specific combination of powers of \(a\) and \(b\), multiplied by a binomial coefficient.

In this problem, we are expanding \( (3x-2)^9 \). The expansion will have a total of \(n+1\) terms, which means there will be 10 terms (since \(n = 9\)). Understanding how to expand such expressions is key to applying the Binomial Theorem effectively.

By following the step-by-step process, we can systematically find any term within the expansion without having to expand the whole expression.

binomial coefficient

The binomial coefficient \( \binom{n}{k} \) is a central part of the Binomial Theorem. It represents the number of ways to choose \(k\) elements from a set of \(n\) elements and is defined as:

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

For our exercise, we needed the binomial coefficient for the 3rd term in the expansion of \( (3x-2)^9 \), which means we set \(k = 2\). Using the formula, we calculated \( \binom{9}{2} = 36 \). This coefficient determines how many times each combination of powers of \(a\) and \(b\) should appear in the term.

powers

Each term in the binomial expansion \( (a+b)^n \) has powers of \(a\) and \(b\) that sum to \(n\). For a given term indexed by \(k\), the powers are \((n-k)\) for \(a\) and \(k\) for \(b\).

In our problem, we needed the 3rd term: \(k+1 = 3 \) so \(k = 2\). Therefore, the powers are:

• Power of \(a = (9-2) = 7\)
• Power of \(b = 2\)

Calculating these powers gave us the corresponding values: \( (3x)^7 = 2187x^7 \) and \( (-2)^2 = 4 \).

terms in binomial expansion

Each term in the binomial expansion \((a + b)^n\) takes the form:

\[T_{k+1} = \binom{n}{k} a^{n-k} b^k\]

This structure allows us to systematically find each term. For example, in the 3rd term of the expansion of \((3x-2)^9\), we used:

• \[\binom{9}{2} = 36\]
• \[ (3x)^7 = 2187x^7 \]
• \[ (-2)^2 = 4 \]

Multiplying these parts together, we obtained the term \[T_3 = 36 \times 2187 x^7 \times 4 = 314928 x^7 \].

This step-by-step method ensures accuracy and understanding as it breaks down the complex problem into manageable pieces.