Problem 39 - Algebra for College Students

Question

Expand each binomial and simplify.\(\left(x+\frac{1}{n}\right)^{6}\)

Step-by-Step Solution

Answer

The expanded form is: \( x^6 + \frac{6x^5}{n} + \frac{15x^4}{n^2} + \frac{20x^3}{n^3} + \frac{15x^2}{n^4} + \frac{6x}{n^5} + \frac{1}{n^6} \).

Step 1: Identify the Expression

The expression to expand is \( \left( x + \frac{1}{n} \right)^6 \). This is a binomial raised to the sixth power.

Step 2: Apply the Binomial Theorem

The Binomial Theorem states that \( (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \). Here, \( a = x \), \( b = \frac{1}{n} \), and \( n = 6 \).

Step 3: Expand Using Binomial Coefficients

Calculate each term using the formula: \( \binom{n}{k} x^{n-k} \left(\frac{1}{n}\right)^k \). The coefficients are found using \( \binom{6}{k} \) for \( k = 0 \) to \( 6 \).

Step 4: Calculate Each Term in the Expansion

Evaluate each term:- For \( k = 0 \): \( \binom{6}{0} x^6 \left(\frac{1}{n}\right)^0 = x^6 \)- For \( k = 1 \): \( \binom{6}{1} x^5 \left(\frac{1}{n}\right)^1 = 6x^5 \frac{1}{n} \)- For \( k = 2 \): \( \binom{6}{2} x^4 \left(\frac{1}{n}\right)^2 = 15x^4 \frac{1}{n^2} \)- For \( k = 3 \): \( \binom{6}{3} x^3 \left(\frac{1}{n}\right)^3 = 20x^3 \frac{1}{n^3} \)- For \( k = 4 \): \( \binom{6}{4} x^2 \left(\frac{1}{n}\right)^4 = 15x^2 \frac{1}{n^4} \)- For \( k = 5 \): \( \binom{6}{5} x^1 \left(\frac{1}{n}\right)^5 = 6x \frac{1}{n^5} \)- For \( k = 6 \): \( \binom{6}{6} x^0 \left(\frac{1}{n}\right)^6 = \frac{1}{n^6} \)

Step 5: Combine and Simplify the Expression

Write the expanded expression by combining all terms:\[ x^6 + \frac{6x^5}{n} + \frac{15x^4}{n^2} + \frac{20x^3}{n^3} + \frac{15x^2}{n^4} + \frac{6x}{n^5} + \frac{1}{n^6} \]

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool that helps in expanding expressions of the form \((a + b)^n\). This theorem tells us how to expand binomials without having to multiply them manually many times. It uses binomial coefficients to determine the weights of each term in the expansion.
For example, in the expression \((x + \frac{1}{n})^6\), the binomial theorem allows us to write it as a sum of terms based on the formula \[ (a+b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]The letters \(a\) and \(b\) are the parts of the binomial, and \(n\) is the power.

• \(\binom{n}{k}\) indicates the binomial coefficient for each term.
• These coefficients indicate how many ways you can pick \(k\) elements out of \(n\), hence determining the number of combinations.
• Each term in the expanded polynomial corresponds to a different value of \(k\), starting from \(0\) up to \(n\).

In our exercise, the variables are \(a = x\), \(b = \frac{1}{n}\), and \(n = 6\). By applying the binomial theorem, the expansion directly follows using simple calculations for each of the terms.

Pascal's Triangle

Pascal’s Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above. It's often used to find the coefficients needed in binomial expansions. The rows of Pascal's Triangle correspond to the powers of binomials, providing a quick way to determine the coefficients.
For example, the 6th row of Pascal's Triangle gives us the coefficients for expanding \((x + \frac{1}{n})^6\). This row is \([1, 6, 15, 20, 15, 6, 1]\), which are the binomial coefficients \(\binom{6}{0}, \binom{6}{1}, \ldots, \binom{6}{6}\).

• The first and last numbers in each row are always 1, as the number of ways to choose all or none from \(n\) is one.
• The symmetry of the triangle means that \(\binom{n}{k} = \binom{n}{n-k}\).

Pascal's Triangle not only simplifies computation of coefficients but also aids visual understanding of how these coefficients are derived and applied in expansions.

Algebraic Expressions

An algebraic expression is a combination of numbers, variables, and arithmetic operations. In the context of binomial expansion, expressions involve variables raised to powers, summed, and multiplied by coefficients.
Understanding algebraic expressions involves recognizing the structure of each term, which is crucial in expansions like \((x + \frac{1}{n})^6\). Here, each term in the expanded expression is a product of a coefficient and powers of the variable \(x\) and constant \(\frac{1}{n}\), which resembles a polynomial with multiple terms where each term decreases in degree.

• Each term in the expression \(x^6 + \frac{6x^5}{n} + \frac{15x^4}{n^2} + \cdots\) is made up by multiplying a coefficient by powers of \(x\) and \(\frac{1}{n}\)
• The expression is simplified by ensuring all terms are combined properly, following algebraic rules.

Effective manipulation of algebraic expressions is foundational in solving and simplifying binomial expansions, being essential for both practical computations and deeper understanding.