Problem 46 - Discrete Mathematics with Applications

Question

Verify each. $$(n+1) !+n !=(n+2) n !$$

Step-by-Step Solution

Answer

To verify the given equation \((n+1)! + n! = (n+2)n!\), we first expand the factorials as \((n+1)n! + n! = (n+2)n!\). Then, factor out the common term \(n!\) on both sides, giving us \((n+1) + 1 = n+2\). Since both sides are equal, the equation is verified.

Step 1: Expand both sides of the equation

Write out the factorials (n+1)! and n! on the left side and the (n+2)n! on the right side of the equation. Recall the definition of a factorial: n! = n(n-1)(n-2)...(3)(2)(1). We can apply this to the given equation as follows: \((n+1)! + n! = (n+2)n!\) Step 2: Simplify the equation

Step 2: Simplify the equation

Next, we will simplify the equation by factoring out the common term n! from both sides of the equation. Remember that: \((n+1)! = (n+1)n(n-1)(n-2)...(3)(2)(1) = (n+1)n!\) So, we rewrite the given equation as: \((n+1)n! + n! = (n+2)n!\) Now, factor n! from the left side: {n!(n+1) + n! = (n+2)n!} Step 3: Solve for n!

Step 3: Solve for n!

To make the equation clearer, let's isolate n! in the equation for better comparison: \((n+1)n! + n! = (n+2)n!\) Divide both sides by n!: \(n+1 + 1 = n+2\) Step 4: Verify the simplified equality

Step 4: Verify the simplified equality

The equation now looks like this: \(n+1 + 1 = n+2\) Which simplifies to: \(n+2 = n+2\) Clearly, both sides of the equation are equal, so the given equation is verified: \((n+1)! + n! = (n+2)n!\)

Key Concepts

Factorial Equation

A factorial, denoted with an exclamation mark (e.g., \( n! \)), represents the product of all positive integers up to a specified number. For example, \(4! = 4 \times 3 \times 2 \times 1 = 24\). The factorial equation in our problem, \((n+1)! + n! = (n+2)n!\), involves comparing two expressions by expanding and simplifying factorials.
Understanding how to manipulate factorials is crucial in solving factorial equations. Start by expanding the factorial terms. For instance, \((n+1)!\) is expanded to \((n+1) \cdot n!\), and \((n+2)n!\) already shows a clear multiplicative structure. Breaking down each factorial into a product of integers simplifies the equation, making it easier to observe common factors.
By recognizing common terms and factoring them out, you can often simplify even complex-looking factorial equations into simple statements. This kind of problem-solving technique is invaluable in mathematical computations involving factorials.

Mathematical Proof

Proving mathematical equations involves ensuring that two expressions are equivalent under given operations and constraints. In our context, this means proving the equality \((n+1)! + n! = (n+2)n!\) by checking both sides of the equation.

• Expand Factorials: Use the definition of factorials to break down terms.
• Factor Common Terms: Look for constants or variables that appear in more than one term, allowing simplification.
• Verify Simplification: Once simplified, ensure the new equation is equivalent to the original, avoiding errors during manipulation.

Successfully proving an equation like this not only verifies the problem but strengthens understanding of how equations can be simplified and compared, especially when dealing with factorials - a foundation for more complex mathematical concepts.

Algebraic Manipulation

Algebraic manipulation is a key tool used to rearrange, simplify, or factor algebraic expressions or equations to make them more understandable or to solve them. In this exercise's solution, algebraic manipulation was utilized extensively to simplify both sides of the equation.

• Algebraic manipulation often involves factoring, expanding, and simplifying expressions.
• By taking \(n!\) as a common term from both sides of the equation, the problem became considerably simpler.
• Strategically using addition, subtraction, multiplication, and division, we can achieve simplification or find equivalent expressions.

The ultimate goal is not just to simplify the equation but to reach a form where all terms have been accounted for and balance is achieved on both sides. This step-by-step approach demonstrates how effective algebraic manipulation can drastically change the complexity of a mathematical problem.