Problem 23 - Algebra and Trigonometry

Question

Evaluate the expression. $$ \left(\begin{array}{l}{5} \\ {0}\end{array}\right)+\left(\begin{array}{l}{5} \\\ {1}\end{array}\right)+\left(\begin{array}{l}{5} \\\ {2}\end{array}\right)+\left(\begin{array}{l}{5} \\\ {3}\end{array}\right)+\left(\begin{array}{l}{5} \\\ {4}\end{array}\right)+\left(\begin{array}{l}{5} \\ {5}\end{array}\right) $$

Step-by-Step Solution

Answer

The sum is 32.

Step 1: Understanding the Problem

We need to evaluate the sum of several binomial coefficients: \( \binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} \). Binomial coefficients are calculated using the formula: \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \) where \( n! \) (n factorial) is the product of all positive integers up to \( n \).

Step 2: Calculate Individual Binomial Coefficients

We will calculate each binomial coefficient separately.1. \( \binom{5}{0} = 1 \) because \( \frac{5!}{0!5!} = 1 \).2. \( \binom{5}{1} = 5 \) because \( \frac{5!}{1!4!} = 5 \).3. \( \binom{5}{2} = 10 \) because \( \frac{5!}{2!3!} = 10 \).4. \( \binom{5}{3} = 10 \) because \( \frac{5!}{3!2!} = 10 \).5. \( \binom{5}{4} = 5 \) because \( \frac{5!}{4!1!} = 5 \).6. \( \binom{5}{5} = 1 \) because \( \frac{5!}{5!0!} = 1 \).

Step 3: Sum the Coefficients

Add the results of the individual binomial coefficients calculated in the previous step:\[ 1 + 5 + 10 + 10 + 5 + 1 = 32 \]

Step 4: Verify Using Binomial Theorem

According to the binomial theorem, \((a + b)^n\) expands to the sum: \[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \].For \(n = 5\), \((1+1)^5 = 2^5 = 32\). Thus, \( \binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 32 \), confirming our calculation.

Key Concepts

Binomial Theorem

The binomial theorem is a foundational concept in algebra that provides a formula for expanding expressions raised to a power. This theorem states that for any positive integer \( n \), the expression \((a + b)^n\) can be expanded as: \[ \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \] where \( \binom{n}{k} \) are the binomial coefficients. Each term in the expansion represents a different way to choose \( k \) items out of \( n \) and is weighted by the factor \( a^{n-k}b^k \).
In the context of our exercise, the summation of binomial coefficients for \( n=5 \) and \( a = b = 1 \) simplifies to \( (1+1)^5 \), which equals 32. This reveals a fascinating connection between binomial coefficients and powers of 2, particularly that the sum of binomial coefficients equals \( 2^n \) for \( n \). It effectively proves the accuracy of manually summing these coefficients.

Factorial

Factorials play an essential role in calculating binomial coefficients. The factorial of a number \( n \) is represented by \( n! \) and is defined as the product of all positive integers from 1 to \( n \). For instance, \( 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \).
In binomial coefficients, factorials help determine combinations, as shown in the formula \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \). Each part of this formula relates the set-up of arrangements for \( n \) items taken \( k \) at a time without considering order.
Understanding factorials is crucial to performing these calculations since they are the backbone of determining the number of possible ways to choose \( k \) elements from \( n \) elements.

Combinatorics

Combinatorics is the branch of mathematics focused on counting, arrangement, and combination of sets of elements. It's about determining the number of ways certain things can occur. In the context of our exercise, combinatorics gives meaning to binomial coefficients, which represents combinations.
When calculating \( \binom{5}{3} \), for example, it essentially means "how many different ways can you choose 3 objects from a set of 5?" This pulls from fundamental combinatorial principles.
This core idea supports much of higher-level mathematics and has real-world applications, such as creating combinations for passwords or organizing teams in sports. By leveraging the tools of combinatorics, we gain systematic approaches to solving problems involving arrangements and selections.

Arithmetic Summation

Arithmetic summation involves the addition of a sequence of numbers, and it helps simplify and work through series or sequence-related problems. In our exercise, summing individual binomial coefficients \( \binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} \) is an example of arithmetic summation.
We calculate each coefficient and add these values systematically:

• \( \binom{5}{0} = 1 \)
• \( \binom{5}{1} = 5 \)
• \( \binom{5}{2} = 10 \)
• \( \binom{5}{3} = 10 \)
• \( \binom{5}{4} = 5 \)
• \( \binom{5}{5} = 1 \)

The sum is \( 1 + 5 + 10 + 10 + 5 + 1 = 32 \). These steps and processes of summation can be applied across various mathematical fields, and they underline the importance of understanding basic arithmetic skills in algebra and number theory.

Mathematical Proof

Mathematical proof is a careful demonstration showing the truth of a given proposition or formula. Proofs are essential in mathematics because they confirm that our results are universally applicable. In our exercise, we verified that the sum of the binomial coefficients equals \( (1+1)^5 = 2^5 = 32 \) by appealing to the binomial theorem.
The process of verifying through the binomial theorem serves as a proof for our manual calculations of binomial coefficients. Using a different perspective via the binomial theorem demonstrates that each part binds together logically to final results.
Proofs often help in exploring deeper relationships between mathematical concepts, offering concrete reasoning behind the formulas and results we use daily. This validation ensures that calculations are consistent and accurate across various applications and contexts in both theoretical and practical scenarios.