Problem 36 - A Graphical Approach to College Algebra

Question

Write the binomial expansion for each expression. $$\left(3+\frac{y}{3}\right)^{5}$$

Step-by-Step Solution

Answer

The binomial expansion is \(243 + 135y + 30y^2 + 10y^3 + \frac{5}{27}y^4 + \frac{1}{243}y^5\).

Step 1: Identify the Components of the Binomial

The expression is \((3 + \frac{y}{3})^5\). This is a binomial expression with terms \(a = 3\) and \(b = \frac{y}{3}\), and exponent \(n = 5\).

Step 2: Apply the Binomial Theorem

The Binomial Theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). For our expression, substitute \(a = 3\), \(b = \frac{y}{3}\), and \(n = 5\) into the formula.

Step 3: Calculate Binomial Coefficients

For \(k = 0, 1, 2, 3, 4, 5\), calculate the binomial coefficients \(\binom{5}{k}\):- \(\binom{5}{0} = 1\)- \(\binom{5}{1} = 5\)- \(\binom{5}{2} = 10\)- \(\binom{5}{3} = 10\)- \(\binom{5}{4} = 5\)- \(\binom{5}{5} = 1\)

Step 4: Expand the Binomial

Using the coefficients and formula, expand:\[(3 + \frac{y}{3})^5 = \sum_{k=0}^{5} \binom{5}{k} (3)^{5-k} \left(\frac{y}{3}\right)^k\]

Step 5: Compute Each Term

Calculate each term in the expansion:- \(k=0\): \(\binom{5}{0} \cdot 3^{5-0} \cdot \left(\frac{y}{3}\right)^0 = 243\)- \(k=1\): \(\binom{5}{1} \cdot 3^{5-1} \cdot \left(\frac{y}{3}\right)^1 = 405y/3 = 135y\)- \(k=2\): \(\binom{5}{2} \cdot 3^{5-2} \cdot \left(\frac{y}{3}\right)^2 = 270y^2/9 = 30y^2\)- \(k=3\): \(\binom{5}{3} \cdot 3^{5-3} \cdot \left(\frac{y}{3}\right)^3 = 90y^3/27 = 10y^3\)- \(k=4\): \(\binom{5}{4} \cdot 3^{5-4} \cdot \left(\frac{y}{3}\right)^4 = 15y^4/81 = \frac{5y^4}{27}\)- \(k=5\): \(\binom{5}{5} \cdot 3^{5-5} \cdot \left(\frac{y}{3}\right)^5 = y^5/243 = \frac{y^5}{243}\)

Step 6: Combine the Terms

Add all the terms together to get the final expansion:\[243 + 135y + 30y^2 + 10y^3 + \frac{5}{27}y^4 + \frac{1}{243}y^5\]

Key Concepts

Binomial Theorem

The **Binomial Theorem** is an excellent tool in algebra for expanding expressions of the form \((a+b)^n\). It simplifies the expansion by using patterns formed in pascal’s triangle and binomial coefficients. In the expression \((a+b)^n\), \(a\) and \(b\) are the terms you want to expand, and \(n\) indicates the power.
Here’s how you apply it: the theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). The symbol \( \binom{n}{k} \) denotes binomial coefficients, which can be found through combinatorics.
The magic of the binomial theorem lies in these coefficients. They tell us how many ways to choose \(k\) items from \(n\) items without caring about the order. Different combinations give us different coefficients.
Using this theorem with our example, \((3 + \frac{y}{3})^5\), we identify \(a = 3\), \(b = \frac{y}{3}\), and \(n = 5\). Substitute these values into the formula to expand step by step.
Each term of this expansion is a blend of powers of \(a\) and \(b\), multiplied by their corresponding binomial coefficient. It allows mathematicians and students alike to effortlessly expand binomials, which in turn aids deeper algebraic manipulations.

algebra

**Algebra** is the field of mathematics that uses symbols and letters to represent numbers and quantities in formulas and equations. It's like a universal language for expressing mathematical ideas. This language helps us solve equations and understand patterns in numbers.
In the context of binomial expansion, algebra plays a crucial role in simplifying terms and making calculations easier. For instance, understanding how to raise a number to a power, multiply expressions, and apply distributive properties are all algebraic concepts.
In the given problem, \((3 + \frac{y}{3})^5\), algebraic skills are useful for manipulating fractions and whole numbers, allowing each term to be computed separately. After computing, you add the terms to get the final expanded form.
Algebra helps you identify patterns and systematically solve equations. Recognizing these patterns can significantly simplify a mathematician's work. This also allows for greater insight into complex problems and generalizations in mathematics.

combinatorics

**Combinatorics** is a branch of mathematics focused on counting, arrangement, and combination of objects. It provides the foundation for calculating the binomial coefficients found within the binomial theorem.
These coefficients, represented as \( \binom{n}{k} \), show the number of ways to choose \(k\) elements from a total of \(n\) elements. They arise from a fundamental part of combinatorial analysis—choosing subsets. This is essential in the binomial expansion process.
In our problem, the coefficients \( \binom{5}{k} \) are calculated for each \(k\) from 0 to 5.

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

These coefficients directly influence the terms in the expansion, making combinatorics a powerful tool in algebraic expressions like this one. Understanding them not only boosts your mathematical skills but also opens doors to solving complex combinatorial problems.