Problem 22 - Precalculus: Mathematical for Calculus

Question

Use the Binomial Theorcm to expand the expression. $$(1-x)^{5}$$

Step-by-Step Solution

Answer

\(1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5\).

Step 1: Understand the Binomial Theorem

The Binomial Theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). In this problem, set \(a = 1\), \(b = -x\), and \(n = 5\).

Step 2: Identify the General Term

The general term in the expansion \(T_k\) is given by \( \binom{5}{k} a^{5-k} b^k \), which becomes \( \binom{5}{k} (1)^{5-k} (-x)^k \).

Step 3: Expand Using the Binomial Coefficients

Calculate each term for \(k = 0\) to \(5\):* For \(k = 0\): \(\binom{5}{0}(1)^5(-x)^0 = 1\)* For \(k = 1\): \(\binom{5}{1}(1)^4(-x)^1 = -5x\)* For \(k = 2\): \(\binom{5}{2}(1)^3(-x)^2 = 10x^2\)* For \(k = 3\): \(\binom{5}{3}(1)^2(-x)^3 = -10x^3\)* For \(k = 4\): \(\binom{5}{4}(1)^1(-x)^4 = 5x^4\)* For \(k = 5\): \(\binom{5}{5}(1)^0(-x)^5 = -x^5\)

Step 4: Combine All the Terms

Sum all terms from previous step to get the expanded form: \(1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5\).

Key Concepts

Understanding Binomial Coefficients

Binomial coefficients are essential components of the binomial theorem. They are denoted by \( \binom{n}{k} \), pronounced "n choose k". These coefficients represent the number of ways to choose \( k \) elements from a set of \( n \) elements without regard to order.

In the context of the binomial theorem, binomial coefficients help in locating the position of each term in the expansion of a binomial expression. For instance, if you have an expression like \((1 - x)^5\), each term in the expansion will involve a binomial coefficient which informs how many "1s" and "-xs" are being multiplied.

To calculate a binomial coefficient \( \binom{n}{k} \), use the formula:

\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]

Here, \( ! \) denotes factorial, which is the product of all positive integers up to a given number. For example, if \( n = 5 \) and \( k = 2 \), you calculate \( \binom{5}{2} \) as follows:

• \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \)
• \( 2! = 2 \times 1 = 2 \)
• \( 3! = 3 \times 2 \times 1 = 6 \)
• Therefore, \( \binom{5}{2} = \frac{120}{2 \times 6} = 10 \)

This coefficient helps determine the coefficient of each term in the polynomial expansion.

Polynomial Expansion Using the Binomial Theorem

Polynomial expansion simplifies expressions raised to a power, using the binomial theorem, helping us see each component of the expansion. Consider the expression \((1-x)^5\). Using the binomial theorem, it is expanded according to:

\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]

In our case, set \( a = 1 \), \( b = -x \), and \( n = 5 \). To find each term of the expansion, apply:

• \( \binom{5}{k} (1)^{5-k} (-x)^k \) for each \( k \)

For \( k = 0 \) through \( 5 \), calculate each term:

• First term: \( \binom{5}{0}(1)^5(-x)^0 = 1 \)
• Second term: \( \binom{5}{1}(1)^4(-x)^1 = -5x \)
• Third term: \( \binom{5}{2}(1)^3(-x)^2 = 10x^2 \)
• Fourth term: \( \binom{5}{3}(1)^2(-x)^3 = -10x^3 \)
• Fifth term: \( \binom{5}{4}(1)^1(-x)^4 = 5x^4 \)
• Sixth term: \( \binom{5}{5}(1)^0(-x)^5 = -x^5 \)

Combining all these, the expanded form of the polynomial is:

\[ 1 - 5x + 10x^2 - 10x^3 + 5x^4 - x^5 \]

The Role of Combinatorics in Polynomial Expansions

Combinatorics is a branch of mathematics dealing with the counting, arrangement, and combination of objects. In polynomial expansions, combinatorics provides the theoretical foundation for understanding how terms are selected and combined.

The concepts of combinations and binomial coefficients are the heart of combinatorics’ contribution to binomial expansions. When faced with an expression like \((1-x)^5\), combinatorics explains how the terms \(1\) and \(-x\) are combined to form different power terms.

In the binomial expansion, each term represents a combination of choosing \( k \) instances of \(-x\) from a total of 5 terms. The binomial coefficient \( \binom{5}{k} \) is thus a combinatorial number indicating how many ways this choice is possible.

This perspective helps to ensure that all terms in the expansion, from the constant term \(1\) to the highest degree term \(-x^5\), are correctly and systematically derived. Therefore, combinatorics is not just counting; it's an indispensable tool for systematically organizing terms and understanding the underlying logic in polynomial expansions.