Problem 12 - Algebra and Trigonometry

Question

Use mathematical induction to prove that the formula is true for all natural numbers \(n\). $$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\cdots+\frac{1}{n(n+1)}=\frac{n}{(n+1)} $$

Step-by-Step Solution

Answer

The formula \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1} \) is true for all natural numbers.

Step 1 - Base Case

We need to show that the formula \( \frac{1}{1 \cdot 2} = \frac{1}{2} = \frac{1}{1+1} \) holds for \( n = 1 \). Clearly, the formula is true for this base case.

Step 2 - Inductive Hypothesis

Assume that the formula is true for some natural number \( k \), so \[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1}. \] We will use this hypothesis to prove that the formula holds for \( k+1 \).

Step 3 - Inductive Step

To prove the formula for \( n = k+1 \), consider the expression: \[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)}. \] Using the inductive hypothesis, substitute \( \frac{k}{k+1} \) for the first \( k \) terms, giving us: \[ \frac{k}{k+1} + \frac{1}{(k+1)(k+2)}. \]

Step 4 - Simplification

Combine the fractions: \[ \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)+1}{(k+1)(k+2)}. \] Simplifying the numerator gives \( k(k+2) + 1 = k^2 + 2k + 1 \), which factors to \((k+1)^2\). Therefore, \[ \frac{k(k+2) + 1}{(k+1)(k+2)} = \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}. \]

Step 5 - Conclusion

We have shown that \( \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2} \). By mathematical induction, the original formula holds true for all natural numbers \( n \).

Key Concepts

Natural Numbers

In mathematics, natural numbers form the set of positive integers beginning from 1. They are the numbers we typically use for counting everyday items, like apples or books, making them simple yet fundamental.
Natural numbers are important in proofs, especially in mathematical induction, as they provide a clear starting point and increment naturally.
The sequence of natural numbers is infinite: 1, 2, 3, and so on. - They do not include zero or negative numbers, unless specified in certain contexts. - Commonly used to explain and understand basic arithmetic and advanced concepts. Their role in induction proofs is crucial because the principle of induction relies on the well-ordering of these numbers, allowing us to verify a statement for all values in this infinite set.

Inductive Hypothesis

The inductive hypothesis is a key assumption in the process of mathematical induction. When you execute an induction proof, you assume that the statement holds for a particular natural number, often represented as 'k'.
For the problem at hand, the inductive hypothesis assumes the formula holds for any arbitrary number 'k'. It looks like this:\[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1}. \]Why is this hypothesis significant? It acts as a bridge that links the base case to all subsequent cases in the proof.- If you successfully prove that the hypothesis leads to the truth of the next case ('k + 1'), the induction step is complete.- It's like saying, if the formula works for 'k', then it must work for 'k + 1'. By allowing this assumption, you establish the ground for proving an infinitely extended logical argument, freeing you from having to prove each case individually.

Base Case

The base case is the initial step in a proof by mathematical induction. It is crucial because it establishes the foundation of the induction process.
In this exercise, the base case is for the smallest natural number, which is 1:\[ \frac{1}{1 \cdot 2} = \frac{1}{2} = \frac{1}{1+1}. \]By showing that the formula is true in this simplest scenario, you:- Prove the starting point of the induction process. - Ensure that the conclusion drawn from the inductive hypothesis holds at the very beginning of the sequence.The base case confirms that the structure we use for mathematical induction is built on solid ground. By proving this step, we make sure the entire induction proof has a valid origin. Without it, the domino effect needed to prove all natural numbers fails to initiate.

Inductive Step

The inductive step is where the power of mathematical induction truly shines. After establishing the truth for the base case and assuming it works for some 'k' (inductive hypothesis), you need to show it holds for 'k + 1'.
In our example, this involves adding the term \( \frac{1}{(k+1)(k+2)} \) to the assumption:\[ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} \]Using the hypothesis, we combine it as:\[ \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2)+1}{(k+1)(k+2)} \]Simplify the numerator to \( (k+1)^2 \), which factors easily, and the left side reduces to the right side:\[ \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2}. \]The inductive step strikes through:- Verifying the leap from case 'k' to 'k + 1'. - Demonstrating that if it works for one, it must work for the next.Thus, mathematical induction proves the statement for all natural numbers, one step at a time.