Problem 37 - College Algebra with Corequisite Support

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 9th term is \(1082565 a^3 b^{16}\).

Step 1: Identify the binomial expression and term

The binomial expression is \((a - 3b^2)^{11}\). We need to find the 9th term of its expansion. In a binomial expansion, the terms are given by the Binomial Theorem.

Step 2: Apply the Binomial Theorem

The Binomial Theorem states: \((x + y)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k\). For the binomial \((a - 3b^2)^{11}\), this becomes \((a + (-3b^2))^{11}\). The general term (\(k\)-th term) is given by \(\binom{11}{k} a^{11-k} (-3b^2)^k\).

Step 3: Determine the specific term by setting \(k\)

The term numbers start from 0 in the Binomial Theorem expansion. To find the 9th term, we set \(k = 8\), since the 9th term is actually the term with \(k=8\) in this series (considering \(k+1\)-th term is required for 9th position).

Step 4: Calculate the binomial coefficient

The binomial coefficient is \(\binom{11}{8}\). Calculate it as follows: \(\binom{11}{8} = \binom{11}{3}\) because of symmetry, which gives \(\frac{11 \cdot 10 \cdot 9}{3 \cdot 2 \cdot 1} = 165\).

Step 5: Compute the powers of the terms

Calculate \(a^{11-8} = a^3\) and \((-3b^2)^8 = (3^8)b^{16}\). Since \(3^8 = 6561\), the expression becomes \(6561b^{16}\).

Step 6: Combine the results to identify the term

Combine the calculated values to form the term: \(T_9 = 165 \times a^3 \times 6561 \times b^{16}\). Thus, \(T_9 = 165 \times 6561 \times a^3 \times b^{16}\).

Step 7: Compute the final term

Multiply the numerical coefficients: \(165 \times 6561 = 1082565\). Thus, the 9th term is \(1082565 a^3 b^{16}\).

Key Concepts

Binomial Expansion

The Binomial Expansion allows us to express any power of a binomial, which is an expression involving exactly two terms, raised to a positive integer power.
Instead of manually multiplying the binomial factors, the Binomial Theorem provides a systematic way to expand such expressions. It enables you to find terms without fully expanding the polynomial. Given a binomial expression of the form \[ (x + y)^n, \]we can expand it using \[ \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k. \]This expansion includes a sum of terms, and each term is a product of powers of the binomial's components. This significantly simplifies the process when dealing with large powers.

• The starting power of the first term decreases with each subsequent term, while the starting power of the second component increases.
• We use a summation notation, making it clear that you add up all the individual terms to achieve the full expansion.

This approach is particularly useful in algebra for finding specific terms, like the ninth term in our given problem, without expanding the entire sequence.

Binomial Coefficient

Each term in the binomial expansion is determined not just by the powers of the binomial components but also by the binomial coefficients, represented by \[ \binom{n}{k}. \]These coefficients are pivotal because they determine the weight or multiplier of each term in the expansion. To determine a binomial coefficient, we use the formula: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!}. \]This formula utilizes the factorial function, which is a shorthand way to multiply a series of descending positive integers, making it a vital component of combinatorial mathematics.Symmetry
An interesting feature of binomial coefficients is their symmetry. Specifically, \[ \binom{n}{k} = \binom{n}{n-k}. \]This property is helpful in simplifying computations, just like in the exercise where we simplified \[ \binom{11}{8} \] by computing \[ \binom{11}{3} \] instead, which is much easier.Understanding binomial coefficients not only aids in simplifying expansions but also helps in calculating probabilities, creating algebraic proofs, and identifying patterns in mathematical problems.

Polynomial Terms

In the context of binomial expansion, each term in the polynomial can be thought of as a distinct unit with a specific combination of components and numerical coefficients. These terms are crucial in understanding the expanded form of a binomial expression.Understanding Terms in Expansion
Each term follows the general form \[ \binom{n}{k} x^{n-k} y^k, \] where x and y are the elements of the original binomial, and n is the total power. For instance, in our exercise's binomial \[ (a - 3b^2)^{11}, \] we are particularly interested in constructing a specific term, the ninth term. This is done by calculating the powers and associated coefficients for the terms \[ a^{11-k} \] and \[ (-3b^2)^k. \]

• The exponent of a seen in \[ a^{11-k} \] comes from reducing through each term as we move across the expansion.
• The exponent of \[ (-3b^2)^k \] increases with each term, introducing a negative sign or not depending on the odd/even status of k.

Combining these powers gives us the distinct terms in the polynomial. Additionally, the numerical coefficients are determined by multiplying the binomial coefficient and the result of the calculated power of the binomial components, ensuring the entire term's value is precise and complete.