Problem 17 - Algebra and Trigonometry

Question

Expand each expression using the Binomial Theorem. $$ (x+1)^{5} $$

Step-by-Step Solution

Answer

(x+1)^5 = x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1

Step 1: Write the Binomial Theorem Formula

The Binomial Theorem states: \[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] where \(a = x\), \(b = 1\), and \(n = 5\).

Step 2: Identify the Binomial Coefficient

The binomial coefficient \(\binom{n}{k}\) is calculated using: \[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \] For each term, replace \(n = 5\) and calculate for \(k = 0, 1, 2, 3, 4, 5\).

Step 3: Expand Each Term

For each value of \(k\) from 0 to 5, compute: \[ \binom{5}{k} x^{5-k} 1^{k} \] Calculate and write each term.

Step 4: Combine the Terms

Sum the terms to get the expanded form: \[ (x+1)^5 = \binom{5}{0} x^{5-0} 1^{0} + \binom{5}{1} x^{5-1} 1^{1} + \binom{5}{2} x^{5-2} 1^{2} + \binom{5}{3} x^{5-3} 1^{3} + \binom{5}{4} x^{5-4} 1^{4} + \binom{5}{5} x^{5-5} 1^{5} \]

Step 5: Calculate Each Coefficient and Simplify

Compute the binomial coefficients and simplify each term: \[ (x+1)^5 = 1x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 \]

Key Concepts

Binomial Expansion

The binomial expansion involves expanding an expression raised to a power. When we have expressions like \( (x + 1)^5 \), instead of multiplying \( (x + 1) \) repeatedly, we can use the Binomial Theorem. The Binomial Theorem helps simplify this process using a formula. This method saves time and reduces errors as compared to manual multiplications. The Binomial Theorem formula is \[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]. This formula involves terms \(a\) and \(b\) raised to varying powers, starting from 0 up to \(n\). Each term in the expansion has a coefficient, which we call the binomial coefficient. The binomial expansion reveals the polynomial that the initial binomial expression corresponds to.

Binomial Coefficient

Binomial coefficients are the numerical factors that multiply the terms in the binomial expansion. These coefficients are represented as \ \binom{n}{k} \ \, which is read as 'n choose k.' The formula to compute a binomial coefficient is: \ \binom{n}{k} = \frac{n!}{k!(n-k)!} \ \. Factorial, denoted by \(\!\) (e.g., 5!), is a product of an integer and all the integers below it. For example, \ 5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1 = 120. \ The binomial coefficient decides how many combinations of k items can be chosen from n items. It determines the weight or importance of each term in the expansion. For instance, in the expansion of \( (x+1)^5 \), the binomial coefficients are calculated for \( k = 0 \) through \( k = 5 \), giving coefficients 1, 5, 10, 10, 5, and 1.

Polynomial Expansion

Polynomial expansion involves expressing a binomial like \ (x + 1)/// raised to a power as a polynomial. Polynomials are algebraic expressions with multiple terms, each consisting of variables and coefficients. The expanded form of a polynomial is broken down into individual terms. Using the Binomial Theorem, the polynomial form of \( (x + 1)^5 \) is derived by combining terms: \ 1x^5 + 5x^4 + 10x^3 + 10x^2 + 5x + 1 \. Each term in this polynomial is made up of:

• The binomial coefficient
• The variable \ x\ raised to descending powers
• Another base, like 1, raised to ascending powers

Polynomial expansion is crucial in algebra as it helps in understanding polynomial functions, their roots, and other related properties.

Combinatorics

Combinatorics is a branch of mathematics dealing with combinations, permutations, and counting. In the context of binomial expansion, it helps in calculating the binomial coefficients. When we use the binomial theorem, we determine how many ways we can choose certain terms from a set, which is a combinatorial problem. The formula for binomial coefficients, \(\binom{n}{k}\), stems from combinatorial principles. Combinatorics explains the underlying principles on how to select terms and how to distribute them within the expansion. It finds applications in many areas, including probability and statistics. For example, the process of choosing how many terms from \ n \ are used in \( (x+1)^5 \) stems from combinatorial reasoning.