Problem 44 - Contemporary Precalculus

Question

Find the indicated term of the expansion of the given expression. fourth, \((a+b)^{6}\)

Step-by-Step Solution

Answer

Answer: The fourth term in the expansion of \((a+b)^{6}\) is \(20a^{3}b^{3}\).

Step 1: Identify the Term We Want to Find

As we want to find the fourth term of the expansion, we set k=3 (since the first term corresponds to k=0, the second term corresponds to k=1, and so on).

Step 2: Apply the Binomial Theorem Formula

Using the binomial theorem formula, we substitute n=6 and k=3: \(T_4 = \binom{6}{3}a^{(6-3)}b^{3}\)

Step 3: Calculate the Combination

Calculate binom{6}{3} by using the formula: \(\binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6!}{3!3!}\) Calculate the factorial values: \(6! = 6 × 5 × 4 × 3 × 2 × 1 = 720\) \(3! = 3 × 2 × 1 = 6\) Now, plug these values into the formula: \(\binom{6}{3} = \frac{720}{(6)(6)} = 20\)

Step 4: Substitute the Combination and Simplify

Substitute the combination value into the formula: \(T_4 = 20a^{3}b^{3}\) The fourth term of the expansion of \((a+b)^6\) is \(20a^{3}b^{3}\).

Key Concepts

Combinatorics

In mathematical terms, combinatorics is the art of counting. This branch helps us determine the number of possible arrangements or selections within a set, without needing to count each possibility individually.
One common aspect of combinatorics that we use in binomial expansions is calculating combinations. Combinations help us figure out how many ways we can choose a certain number of items from a larger group, where the order of the items doesn't matter.

For example, in the exercise, to find the fourth term in the binomial expansion, we calculated the combination \( \binom{6}{3} \). This tells us the number of ways to select 3 elements from a group of 6. The formula for combinations is given by:
\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]
where \( n \) is the total number of items, \( k \) is the number of chosen items, and \(!\) represents a factorial. Calculating combinations efficiently allows us to solve complex problems involving choices and arrangements.

Factorial Calculation

A factorial, denoted with an exclamation mark \(!\), is a product of all positive integers up to a specified number. It plays a crucial role in both permutations and combinations in combinatorics. Factorials help us understand how many ways we can arrange a set of items.

For instance, in our exercise, we calculated the combination by using factorial values. To calculate \( 6! \), we multiply all whole numbers from 6 down to 1:
\[ 6 \times 5 \times 4 \times 3 \times 2 \times 1 = 720 \]
Similarly, for \( 3! \), we compute:
\[ 3 \times 2 \times 1 = 6 \]
These individual factorial values are then used to determine combinations, ensuring a correct count of potential arrangements or selections. Factorials grow very quickly, which is why they're so powerful in combinatorial calculations.

Polynomial Expansion

Polynomial expansion is the process of expressing a polynomial raised to a power as the sum of terms. Each term in the expansion follows a specific pattern defined by the binomial theorem. In simple terms, it breaks down something like \((a+b)^n\) into simpler, expanded terms.

The binomial theorem states that:
\[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
This formula gives us the ability to expand expressions like \((a+b)^6\). The coefficients of each term are determined by the binomial coefficients \( \binom{n}{k} \), and the powers of \(a\) and \(b\) add up to \(n\).

In our specific example, finding the fourth term corresponds to using \(k=3\), which yields the term \(20a^3b^3\) as calculated by plugging into the binomial formula. Understanding polynomial expansion allows us to take a compact expression and convert it to a longer form without manually multiplying it out.