Problem 70 - Algebra 2

Question

What is the third term in the expansion of \((a-b)^{7} ?\) \(\begin{array}{llll}{\text { F. }-21 a^{5} b^{2}} & {\text { G. }-7 a^{6} b} & {\text { H. } 7 a^{6} b} & {\text { J. } 21 a^{5} b^{2}}\end{array}\)

Step-by-Step Solution

Answer

The third term in the expansion of \((a-b)^{7}\) is \(-21*a^{5}*b^{2}\). Therefore the correct answer is F: \(-21*a^{5}*b^{2}\).

Step 1: Binomial Theorem

Firstly, make sure you have a thorough understanding of the Binomial Theorem. Essentially, it provides a way of expanding \((a-b)^{n}\) as a sum of terms in the form \(C(n, k) * a^{n-k} * b^{k}\), where \(C(n, k)\) is the combination of \(n\) items taken \(k\) at a time.

Step 2: Determine the third term

Apply the Binomial Theorem to determine the third term of the expansion. Being asked for the third term means \(k = 2\) in the theorem. The formula then becomes: \(C(7, 2)*a^{7-2}*b^{2}\).

Step 3: Evaluate the third term

Evaluate the combination using \(C(7, 2) = 21\). Therefore, the third term results in \(-21*a^{5}*b^{2}\), because of the negative sign in the binomial.

Key Concepts

Binomial expansion

The Binomial Theorem is a powerful algebraic tool that allows us to expand expressions raised to a power. Expressions in the form of \((a - b)^n\) can be expanded into a series consisting of terms like \(C(n, k) \cdot a^{n-k} \cdot b^k\). Here, \(C(n, k)\) stands for a combination, which we'll explain further on. The benefit of binomial expansion is that it helps break down complex expressions into simpler components, which can be useful in both theoretical and practical scenarios.
When working with binomial expansions, the first thing to do is identify the components, \(a\), \(b\), and \(n\), and then determine how many terms will be involved in the expansion. Each term is generated by changing the value of \(k\) from 0 up to \(n\), which gives us one more term than the power of \(n\).
Knowing how to apply binomial expansion is crucial for solving many algebraic problems, and understanding it deeply enables you to tackle higher-level mathematical concepts with more confidence.

Combination formula

In order to use the Binomial Theorem effectively, understanding the combination formula is essential. The combination formula, denoted as \(C(n, k)\), represents the number of ways to choose \(k\) items from a total of \(n\) items, without regard to the order of selection. The mathematical expression for the combination is given by:
\[ C(n, k) = \frac{n!}{k!(n-k)!} \]
where \(!\) denotes factorial, a product of all positive integers up to a certain number.
For example, to find \(C(7, 2)\), you calculate:

• \(7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1\)
• \(2! = 2 \times 1\)
• \((7-2)! = 5! = 5 \times 4 \times 3 \times 2 \times 1\)

Plug these into the formula to get:
\[ C(7, 2) = \frac{7 \times 6}{2 \times 1} = 21 \]
This result indicates that there are 21 different ways to select 2 items from a group of 7, a calculation that's very handy in binomial expansions.

Polynomial expansion

Polynomial expansion, as facilitated by the Binomial Theorem, is the process of expressing a binomial expression that is raised to some power \(n\) as a sequence of simpler polynomial terms. In our context, the expression \((a-b)^7\) can be broken down into these terms using the combination coefficients, the powers of \(a\), and the powers of \(b\).
The general form of each term is \(C(n, k) \cdot a^{n-k} \cdot b^k\). When expanding \((a-b)^7\), each term’s sign is critical — particularly because of the minus sign in the binomial, which alternatively affects the sign of each term.
When determining a specific term, such as the third term in our exercise, it's crucial to substitute each component accurately. For \(k=2\), it follows:

• \(a^{7-2} = a^5\)
• \(b^2 = b^2\)
• Combination coefficient: \(C(7,2) = 21\)

Putting it all together: \(21 \cdot a^5 \cdot b^2\) becomes the third term \(-21a^5b^2\), accounting for the fact that \(b\) terms bring a negative sign from \((a-b)\) raised to any power. This night be more apparent when working through expansions with many terms, making polynomial expansion a key concept in understanding larger algebraic structures.