Problem 27 - Precalculus Mathematics for Calculus

Question

Use the Binomial Theorem to expand the expression. $$\left(1+\frac{1}{x}\right)^{6}$$

Step-by-Step Solution

Answer

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

Step 1: Write the Binomial Theorem

The Binomial Theorem states that for any positive integer \(n\), \((a+b)^n\) can be expanded as \(\sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). In this expression, \(a = 1\) and \(b = \frac{1}{x}\), and \(n = 6\).

Step 2: Identify the components for expansion

For the expression \(\left(1 + \frac{1}{x}\right)^6\), identify that \(a = 1\), \(b = \frac{1}{x}\), and \(n = 6\). The general term for the Binomial Theorem expansion is \(\binom{n}{k} a^{n-k} b^k\).

Step 3: Calculate each term in the expansion

Calculate the first few terms using the formula: - When \(k=0\), the term is \(\binom{6}{0} (1)^{6} \left(\frac{1}{x}\right)^0 = 1\).- When \(k=1\), the term is \(\binom{6}{1} (1)^{5} \left(\frac{1}{x}\right)^1 = 6\left(\frac{1}{x}\right) = \frac{6}{x}\).- When \(k=2\), the term is \(\binom{6}{2} (1)^{4} \left(\frac{1}{x}\right)^2 = 15\left(\frac{1}{x^2}\right) = \frac{15}{x^2}\).- Continue this pattern until \(k=6\).

Step 4: Continue the expansion for other terms

Continue calculating further terms using the formula: - When \(k=3\), \(\frac{1}{x^3}\), the term is \(\binom{6}{3} (1)^{3} \left(\frac{1}{x}\right)^3 = 20\left(\frac{1}{x^3}\right) = \frac{20}{x^3}\).- When \(k=4\), \(\frac{1}{x^4}\), the term is \(\binom{6}{4} (1)^{2} \left(\frac{1}{x}\right)^4 = 15\left(\frac{1}{x^4}\right) = \frac{15}{x^4}\).- When \(k=5\), \(\frac{1}{x^5}\), the term is \(\binom{6}{5} (1)^{1} \left(\frac{1}{x}\right)^5 = 6\left(\frac{1}{x^5}\right) = \frac{6}{x^5}\).- When \(k=6\), \(\frac{1}{x^6}\), the term is \(\binom{6}{6} (1)^{0} \left(\frac{1}{x}\right)^6 = 1\left(\frac{1}{x^6}\right) = \frac{1}{x^6}\).

Step 5: Compile the full expansion

Putting all the terms together, the expanded form of \(\left(1+\frac{1}{x}\right)^6\) is: \[1 + \frac{6}{x} + \frac{15}{x^2} + \frac{20}{x^3} + \frac{15}{x^4} + \frac{6}{x^5} + \frac{1}{x^6}\]

Key Concepts

Combinatorial Coefficients

Combinatorial coefficients are a fundamental part of the binomial theorem. They are also known as binomial coefficients and are selected from Pascal's triangle, representing ways to choose subsets from a larger set. When expanding the expression \((a + b)^n\), each term includes a coefficient \(\binom{n}{k}\), which is read as "n choose k." This coefficient calculates how many ways you can select \(k\) elements from a total of \(n\) elements.
In the exercise, the use of \(\binom{6}{k}\) for terms in the expression \((1 + \frac{1}{x})^{6}\) shows the application of these coefficients. For example, \(\binom{6}{2}\) computes to 15, meaning there are 15 ways to choose 2 elements from a set of 6. Exploring these coefficients is crucial in understanding how terms are weighted within a polynomial expansion.

Polynomial Expansion

The binomial theorem provides a systematic method for expanding polynomials raised to a power. In essence, an expression like \((a+b)^n\) is converted from a compact form into a sum of multiple terms. Each term in the expansion is derived by the formula \(\binom{n}{k} a^{n-k} b^k\).

• \(a\) and \(b\) are any numbers or expressions.
• \(n\) is a positive integer.

For the given expression \((1 + \frac{1}{x})^6\), setting \(a = 1\) and \(b = \frac{1}{x}\) allows each power of \(\frac{1}{x}\) to appear in sequence. Understanding each stage of expansion helps identify how each term contributes, with the process revealing an increasing complexity and layer to the expression, influencing powers and coefficients closely tied to the binomial theorem.

Rational Expressions

A rational expression is defined as a fraction where both the numerator and the denominator are polynomials. In the binomial theorem context, rational expressions become especially interesting because the addition of terms requires careful handling of fractional forms, often involving rational numbers or variables in denominators.
In \((1 + \frac{1}{x})^6\), the sequence of terms involves fractions like \(\frac{6}{x}\), \(\frac{15}{x^2}\), and so forth, each showcasing decreasing powers of \(x\). Understanding this aspect of the exercise highlights how rational expressions can be simplified or expanded, offering insights into polynomial behavior and the effects of transformations, like inversion in this case, on each term. As the expansion progresses, it's critical to maintain clarity when handling these expressions, ensuring that computation remains precise.