Problem 24 - College Algebra

Question

In Exercises \(11-30,\) use mathematical induction to prove that each statement is true for every positive integer \(n\) $$ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\frac{1}{4 \cdot 5}+\dots+\frac{1}{(n+1)(n+2)}=\frac{n}{2 n+4} $$

Step-by-Step Solution

Answer

The given statement has been shown to be true for every positive integer \(n\) using the principle of mathematical induction.

Step 1: Base Case

First, check the base case i.e. when \(n=1\), the left hand side becomes \(\frac{1}{2\cdot3} = \frac{1}{6}\) and right hand side becomes \(\frac{1}{2\cdot1+4} = \frac{1}{6}\). Hence, the equation holds true for \(n=1\) which is our base case.

Step 2: Inductive Hypothesis

Assume the formula is true for some positive integer \(n\), this step is called the inductive hypothesis. Therefore, it is assumed that \[ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\dots+\frac{1}{(n+1)(n+2)} = \frac{n}{2n+4} \]

Step 3: Inductive Step

Next, show that if the formula is true for \(n\), then it is also true for \(n+1\). Hence, need to prove \[ \frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\dots+\frac{1}{(n+1)(n+2)}+\frac{1}{(n+2)(n+3)}=\frac{n+1}{2(n+1)+4} \] Subtract the original equation assumed to be true from this, to obtain \[ \frac{1}{(n+2)(n+3)} = \frac{n+1}{2(n+1)+4} - \frac{n}{2n+4} \] Simplifying this equation, it can be shown that both sides are equal, thus proving the formula also holds for the case \(n+1\). Therefore, the statement is proven true by mathematical induction.

Key Concepts

Base Case

When using mathematical induction, the first step is to verify the base case. This is an essential step because it confirms that the statement we want to prove is initially true.
In the exercise, we begin by checking the equation when \( n = 1 \).

• The left side of the equation becomes \( \frac{1}{2 \cdot 3} = \frac{1}{6} \).
• The right side also calculates to \( \frac{1}{6} \).

Since both sides are equivalent when \( n = 1 \), the base case holds true. Establishing the base case is like laying the foundation of a building; it must be solid before proceeding to add more levels.

Inductive Hypothesis

The inductive hypothesis is a crucial step where we assume the statement is true for a particular positive integer \( n \). This forms the basis for proving the next step.
The hypothesis might sound like a guess, but it’s more than that—it’s a strategic assumption.
Here’s how it works for this exercise:

• We assume that the equation \( \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{(n+1)(n+2)} = \frac{n}{2n+4} \) holds true for some integer \( n \).

This step doesn’t require proving; rather, it prepares a pathway to show the statement’s continuation for the \( n+1 \) case.

Inductive Step

In the inductive step, we verify that if the statement holds for \( n \), then it must also hold for \( n+1 \). This is, effectively, how we extend the statement's truth to all positive integers.
For this exercise, the goal is to prove:

• \( \frac{1}{2 \cdot 3} + \frac{1}{3 \cdot 4} + \dots + \frac{1}{(n+1)(n+2)} + \frac{1}{(n+2)(n+3)} = \frac{n+1}{2(n+1)+4} \)

By substituting the inductive hypothesis into this equation and simplifying, we demonstrate that both sides remain equal, which is the core of the inductive step. This process confirms that if the equality holds for \( n \), it will also hold for \( n+1 \), sealing a crucial part of the proof.

Proof by Induction

Proof by induction is a powerful technique used in mathematics to show that a statement or formula is true for all natural numbers. It's somewhat akin to dominoes; proving one statement knocks over the next.
This type of proof consists of:

• First establishing the base case.
• Then assuming the statement is true for some \( n \) (inductive hypothesis).
• Lastly, using this assumption to prove the statement holds for \( n+1 \) (inductive step).

If successful, this method confirms the statement for every natural number, beginning at the base case and continuing indefinitely. Understanding proof by induction helps in tackling complex problems by breaking them into approachable steps.