Problem 2 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Expand \(\left(\frac{2}{x}-\frac{x}{2}\right)^{5}\) by the binomial theorem.

Step-by-Step Solution

Answer

The expanded form is: \(\frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}\).

Step 1: Identify the Binomial Expression

The given expression is \( \left( \frac{2}{x} - \frac{x}{2} \right)^5 \). Here, our binomial terms are \( \frac{2}{x} \) and \( -\frac{x}{2} \).

Step 2: Recall the Binomial Theorem

The binomial theorem states that \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\). In our case, \( a = \frac{2}{x} \), \( b = -\frac{x}{2} \), and \( n = 5 \).

Step 3: Compute the Binomial Coefficients

For \( n = 5 \), the coefficients \( \binom{5}{k} \) for \( k = 0, 1, 2, 3, 4, 5 \) are: \( \binom{5}{0} = 1 \), \( \binom{5}{1} = 5 \), \( \binom{5}{2} = 10 \), \( \binom{5}{3} = 10 \), \( \binom{5}{4} = 5 \), and \( \binom{5}{5} = 1 \).

Step 4: Expand the Binomial using the Theorem

Using the formula and substituting the values, we have: \[\left( \frac{2}{x} - \frac{x}{2} \right)^5 = \sum_{k=0}^{5} \binom{5}{k} \left( \frac{2}{x} \right)^{5-k} \left( -\frac{x}{2} \right)^k\]

Step 5: Calculate Each Term in the Expansion

Compute each term from \( k = 0 \) to \( k = 5 \):1. \( k=0 \): \( \binom{5}{0} \left( \frac{2}{x} \right)^5 \left( -\frac{x}{2} \right)^0 = 1 \times \frac{32}{x^5} \times 1 = \frac{32}{x^5} \)2. \( k=1 \): \( \binom{5}{1} \left( \frac{2}{x} \right)^4 \left( -\frac{x}{2} \right)^1 = 5 \times \frac{16}{x^4} \times -\frac{x}{2} = -\frac{40}{x^3} \)3. \( k=2 \): \( \binom{5}{2} \left( \frac{2}{x} \right)^3 \left( -\frac{x}{2} \right)^2 = 10 \times \frac{8}{x^3} \times \frac{x^2}{4} = \frac{20}{x} \)4. \( k=3 \): \( \binom{5}{3} \left( \frac{2}{x} \right)^2 \left( -\frac{x}{2} \right)^3 = 10 \times \frac{4}{x^2} \times -\frac{x^3}{8} = -5x \)5. \( k=4 \): \( \binom{5}{4} \left( \frac{2}{x} \right)^1 \left( -\frac{x}{2} \right)^4 = 5 \times \frac{2}{x} \times \frac{x^4}{16} = \frac{5x^3}{8} \)6. \( k=5 \): \( \binom{5}{5} \left( \frac{2}{x} \right)^0 \left( -\frac{x}{2} \right)^5 = 1 \times 1 \times -\frac{x^5}{32} = -\frac{x^5}{32} \)

Step 6: Write the Final Expanded Expression

Combine all the terms computed in Step 5: \[\frac{32}{x^5} - \frac{40}{x^3} + \frac{20}{x} - 5x + \frac{5x^3}{8} - \frac{x^5}{32}\]

Key Concepts

Binomial Expansion

The binomial expansion is a technique used to expand expressions raised to a power, such as \( (a + b)^n \). This process allows one to break down a power of a sum into a sum of terms, making computations and simplifications much easier. To perform a binomial expansion, the Binomial Theorem comes into play.
The Binomial Theorem states:

• \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

Here, \( n \) represents the power to which the binomial is raised, and \( k \) indicates the position of each term in the expansion. For each \( k\), a binomial coefficient, expressed as \((n \choose k)\), is calculated, followed by evaluating \( a^{n-k} \) and \( b^k \). The sum of these terms results in the complete expanded expression. This method is invaluable, especially as \( n \) becomes large, providing a clear way to find every term's value rather than multiplying manually.
Understanding the strategy for expansion is crucial when simplifying complex algebraic expressions. It provides order and precision to otherwise daunting calculations, converting a single expression into a sum that reveals more insights.

Polynomial Expressions

Polynomial expressions are expressions made up of variables and coefficients, involving only the operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. In the context of binomial expansion, each term in the expanded expression forms part of a larger polynomial.
When we talk of expanding \( (\frac{2}{x} - \frac{x}{2})^5 \), each term like \( \frac{32}{x^5} \) or \( -\frac{40}{x^3} \) appears as part of this polynomial. Each term has a form \( c \cdot x^m\), where \( c\) is the coefficient and \( m\) is the power. Polynomials can be as simple as a single term or consist of hundreds of terms, and they appear in various fields such as engineering, physics, and economics, showcasing the universality of mathematical principles.

• Terms can be constant or involve variables like \( x\).
• Operations include addition, subtraction, and multiplication with real coefficients.

Polynomials help us understand the behavior of functions, as they can represent practically any kind of smooth curve. With polynomial expressions, notably expanded from binomials, each term tells something specific about the equation’s range, intercepts, and overall behavior.

Combinatorial Coefficients

Also known as binomial coefficients, combinatorial coefficients are essential in binomial expansion because they determine the number of ways to choose \( k \) elements from \( n \) elements, and are represented by \( \binom{n}{k} \). This notation provides the numerical factor by which each term in the expansion is multiplied.
Calculating these coefficients efficiently is critical to performing binomial expansions. They are calculated using the formula:

• \( \binom{n}{k} = \frac{n!}{k!(n-k)!} \)

Where \( n! \) is the factorial of \( n \), meaning multiplying \( n \) by all positive integers below it. For example, in our case with \( n = 5 \), we calculate for each \( k\) as:

• \( \binom{5}{0}, \binom{5}{1}, \binom{5}{2}, \ldots, \binom{5}{5} \)

This provides each term's contribution to the final expansion of the polynomial. It emphasizes not just their combinatorial nature but their power in streamlining calculation and simplifying complicated algebraic expressions. This knowledge facilitates smoother expansion and simplification processes, crucially aiding in both academic and real-world problem-solving scenarios.