Problem 37 - College Algebra

Question

For the following exercises, find the indicated term of each binomial without fully expanding the binomial. The ninth term of \(\left(a-3 b^{2}\right)^{11}\)

Step-by-Step Solution

Answer

The ninth term is \(1082565a^3b^{16}\).

Step 1: Identify the Binomial Formula

To find a specific term in the expansion of a binomial expression, use the binomial theorem formula: \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\). This will allow you to compute any specific term in the expansion without fully expanding the expression.

Step 2: Determine the Term's Position

To find the ninth term of the expansion \((a-3b^2)^{11}\), note that the expression follows: \(T_{k+1} = \binom{n}{k} a^{n-k}(-3b^2)^k\). The ninth term is \(T_9\), which corresponds to \(k=8\).

Step 3: Calculate Binomial Coefficient

Calculate the binomial coefficient for the ninth term: \(\binom{11}{8}\). This coefficient can be calculated using the formula \(\binom{n}{k} = \frac{n!}{k!(n-k)!}\). Thus, \(\binom{11}{8} = \frac{11!}{8!3!}\).

Step 4: Simplify the Coefficient Calculation

Simplify \(\frac{11!}{8!3!}\). This is \(\frac{11 \times 10 \times 9}{3 \times 2 \times 1} = 165\). Thus, \(\binom{11}{8} = 165\).

Step 5: Apply the Coefficient and Calculate Variables

Substitute the binomial coefficient and the variables into the term expression: \(T_9 = 165 \cdot a^{11-8} \cdot (-3b^2)^8\).

Step 6: Compute the Resulting Expression

Simplify \(a^3 (-3b^2)^8\) to get \(a^3 \cdot (-3)^8 \cdot (b^2)^8\). Calculate \((-3)^8 = 6561\) and \((b^2)^8 = b^{16}\).

Step 7: Combine All Calculations

Combine the coefficient, power of \(a\), and result from \((-3b^2)^8\). Hence, the ninth term is \(165 \cdot a^3 \cdot 6561 \cdot b^{16} = 1082565a^3b^{16}\).

Key Concepts

Binomial Coefficient

The binomial coefficient is a key concept when working with the binomial theorem. It represents the number of ways to choose a set of items from a larger set, also known as combinations, and is denoted as \( \binom{n}{k} \). In mathematical terms, this is calculated using the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] Here, \( n \) is the total number of items and \( k \) is the number of items to choose. In the context of binomial expansion, the coefficient reflects the weight of each term's contribution to the overall polynomial. In our problem, we calculated \( \binom{11}{8} \) which equaled 165, showing how many ways there are to select the specific terms for the ninth term in the expansion.

• Important in computing terms in binomial expansions.
• Expresses how coefficients change in a polynomial sequence.

Understanding binomial coefficients allows you to determine each term's contribution efficiently without fully expanding the polynomial.

Polynomial Expansion

The binomial theorem provides a method to expand expressions of the form \((x + y)^n\). In our problem, this was used to expand \((a - 3b^2)^{11}\) efficiently. Instead of fully expanding this expression, we focused on finding the specific term directly by using the theorem, which allows jumping to the calculation of any term. Using the expression for the term \( T_{k+1} = \binom{n}{k} x^{n-k} y^k \), we identified the ninth term as involving \( k = 8 \). Polynomial expansion simplifies complex expressions by breaking them down into simpler, individual terms.

• Allows focusing on particular terms in a polynomial.
• Theorem provides general formula for expansion.

The advantage of polynomial expansion is that it gives us a deeper understanding of how each part of the polynomial contributes to the entire expression.

Combinatorics

Combinatorics is a branch of mathematics dealing with combinations, permutations, and counting. In the realm of polynomial expansions, combinatorics is heavily used in calculating binomial coefficients. It provides tools for determining how to arrange or select items. For binomials, using combinatorics means determining how different powers of each variable align in each term of the expansion. In solving our problem, combinatorics principles helped us compute \( \binom{11}{8} \), showing the number of ways variable combinations could be arranged for the ninth term. This validates how each coefficient in a polynomial expansion stems from a combinatorial count.

• Helps in effectively counting permutations and combinations in mathematics.
• Crucial for understanding distributions within polynomial terms.

With a good grasp of combinatorics, you can handle large polynomial expressions with confidence, choosing specific terms without tedious full expansions.