Problem 34 - College Algebra

Question

For the following exercises, find the indicated term of each binomial without fully expanding the binomial. The seventh term of \((a+b)^{11}\)

Step-by-Step Solution

Answer

The seventh term is \(462a^5b^6\).

Step 1: Understanding Binomial Theorem

The binomial theorem states that for any positive integer \( n \), the expansion of \((a + b)^n\) is given by: \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). This means we can find any individual term in the expansion by determining its coefficient and variables of power.

Step 2: Identify Parameters

We need the seventh term of \((a+b)^{11}\). In the binomial theorem, the \(k\)-th term is generally expressed as \(\binom{n}{k-1} a^{n-(k-1)} b^{k-1}\). However, indexing starts from 0, so the seventh term corresponds to \(k = 6\). Thus, we use \(n = 11\) and \(k = 6\).

Step 3: Find Binomial Coefficient

Calculate the binomial coefficient for the seventh term, which is \(\binom{11}{6}\). This is computed as:\[\binom{11}{6} = \frac{11!}{6! \times (11-6)!} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2 \times 1} = 462.\]

Step 4: Determine Powers of Variables

Since we are looking for the term \(k = 6\), from the binomial theorem, we have \(a\) raised to the power of \(11-6 = 5\) and \(b\) raised to the power of \(6\). Therefore, the term is \(a^5 b^6\).

Step 5: Construct the Term

The seventh term of \((a+b)^{11}\) is given by the product of the binomial coefficient \(462\) and the powers of \(a\) and \(b\):\[462 \cdot a^5 \cdot b^6.\]

Key Concepts

Binomial Expansion

The binomial expansion is a powerful algebraic tool used to expand expressions of the form \((a+b)^n\), where \(n\) is a positive integer. This expansion allows us to express the binomial as a sum involving terms of the form \(\binom{n}{k} a^{n-k} b^k\). Each term within the expansion represents one of the infinite cases in which the powers of \(a\) and \(b\) can combine.A key benefit of the binomial expansion is that you do not have to multiply the terms together manually to fully expand the expression. Instead, using the binomial theorem allows you to identify specific terms without needing to calculate the entire expansion.When you use the binomial expansion, each term has a coefficient, which is crucial in calculating the specific term's value. For instance, in our exercise involving \((a+b)^{11}\), the binomial theorem helps find the seventh term directly, without expanding all terms.

Binomial Coefficient

A binomial coefficient, denoted as \(\binom{n}{k}\), represents the number of ways to choose \(k\) items from \(n\) items without regard to order. Binomial coefficients are a significant part of the binomial expansion, serving as the coefficients for each term in the expansion.Mathematically, the binomial coefficient is given by the formula:\[\binom{n}{k} = \frac{n!}{k!(n-k)!}\]where \(!\) represents a factorial, which is the product of an integer and all positive integers below it.For example, in finding the seventh term of \((a+b)^{11}\), we calculate the binomial coefficient for \(k = 6\), resulting in \(\binom{11}{6} = 462\). This coefficient tells us how many times the term \(a^5b^6\) will appear in the expansion.

Powers of Variables

In the binomial expansion, the powers of the variables \(a\) and \(b\) change across each term. This is determined by the binomial theorem and the specific index \(k\) of the term you are considering.Each term is of the form \(\binom{n}{k} a^{n-k} b^k\), where

• \(a\) is raised to the power \(n-k\), and
• \(b\) is raised to the power \(k\).

Understanding these powers is crucial, as they determine the composition and degree of each term in the expansion.For the seventh term in the expansion of \((a+b)^{11}\), the powers of \(a\) and \(b\) are \(11-6=5\) and \(6\), respectively. This results in \(a^5b^6\), coupled with the calculated binomial coefficient to form \(462\cdot a^5\cdot b^6\) as the complete term.