Problem 60 - Discrete Mathematics with Applications

Question

Prove that \(\sum_{i=1}^{n} h_{i}=(n+1) h_{n}-n, n \geq 1\)

Step-by-Step Solution

Answer

Using the principle of mathematical induction, we first prove the base case (n = 1) and then the inductive step (n = k + 1). For the base case, we find that the equation holds true. Then, assuming the equation holds true for n = k, we expand the sum for k + 1 and apply the inductive hypothesis. By rearranging the equation and performing algebraic manipulations, we show that the equation holds true for n = k + 1. Thus, the formula \(\sum_{i=1}^{n} h_{i}=(n+1) h_{n}-n, n \geq 1\) is proven true by the principle of mathematical induction.

Step 1: Base Case (n = 1)

When n = 1, let's show that the equation holds true: \(\sum_{i=1}^{1} h_{i} = (1+1) h_{1} - 1\) On the left-hand side, we have: \(h_1\) On the right-hand side, we have: \((2) h_{1} -1\) Since both expressions are equal, the formula holds true for the base case, n = 1.

Step 2: Inductive Step

Now, let's assume the equation holds for n = k: \(\sum_{i=1}^{k} h_{i} = (k+1) h_{k} - k\) Now we need to show that the proposition also holds for n = k + 1: \(\sum_{i=1}^{k+1} h_{i} = ((k+1)+1) h_{k+1} - (k+1)\)

Step 3: Sum Expansion

First, we expand the sum for k + 1: \(\sum_{i=1}^{k+1} h_{i} = \sum_{i=1}^{k} h_{i} + h_{k+1}\) By the inductive hypothesis, we know that: \(\sum_{i=1}^{k} h_{i} = (k+1) h_{k} - k\) Therefore, we can substitute that into the expanded sum: \((k+1) h_{k} - k + h_{k+1} = (k+2) h_{k+1} - k - 1\) Now, we want to show that the two sides of the equation are equal.

Step 4: Rearrange the Equation

Rearrange the equation to solve for \(h_{k+1}:\) \(h_{k+1} + (k+1) h_{k} - k = (k+2) h_{k+1} - k - 1\) Subtract \(h_{k+1}\) from both sides of the equation: \((k+1) h_{k} - k = (k+1) h_{k+1} - k - 1\) Lastly, add k and subtract \((k+1) h_{k}\) from both sides to show that both sides are equal: \((k+1) h_{k+1} - (k+1) h_{k} = 1\) The equation holds for n = k + 1 if the following is true: \((k+1) (h_{k+1} - h_{k}) = 1\) Since we have shown that the equation holds true for both the base case (n = 1) and the inductive step (n = k + 1), the formula \(\sum_{i=1}^{n} h_{i}=(n+1) h_{n}-n, n \geq 1\) is proven to be true by the principle of mathematical induction.

Key Concepts

Sum of Sequences

Understanding the concept of the "sum of sequences" is essential for solving problems using mathematical induction. When we talk about the sum of a sequence, we are referring to the aggregation of all the terms within that sequence. For any sequence, such as \([ h_1, h_2, ..., h_n ]\), the sum is represented by the notation \(\sum_{i=1}^{n} h_i\), meaning you add from the first term \((h_1)\) to the nth term \((h_n)\).
The key here is to express this sum in a form that reveals a pattern or relationship, which can clarify how the sequence behaves as it progresses. In our exercise, the student needs to prove that \(\sum_{i=1}^{n} h_{i} = (n+1) h_{n} - n\), indicating a fixed relationship that holds for all positive integers. Recognizing how to break down sequences into understandable parts will prepare you for performing induction techniques.

Base Case

The "base case" serves as the starting point in mathematical induction. It involves solving the problem for the smallest value where the statement applies, usually n = 1. If you prove that the formula holds true at this initial step, you create a foundation for further steps.
In the given exercise, we evaluate the base case by setting n = 1 and showing that both sides of the equation \(\sum_{i=1}^{1} h_{i} = (1+1) h_{1} - 1\) produce the same result. For n = 1, the sum simplifies directly to \( h_1\) on the left-hand side. The right-hand side simplifies to \( (2) h_{1} - 1\), which, when calculated, equals \( h_1\) if the initial condition is accurate.
By establishing that the smallest case is correct, you ensure that the logical progression for larger numbers is grounded in validity. Remember, the basis step sets the tone for ensuring the logical flow of mathematical induction.

Inductive Step

The "inductive step" involves proving that if a statement holds for one case, it will also be true for the next case. This portion of the proof allows you to demonstrate continuity of the truth of the statement for all subsequent values.
For the inductive step, you start by assuming the statement is true for an arbitrary positive integer n = k. This assumption is called the "inductive hypothesis." You then apply this assumption to prove the statement for the next integer value, n = k + 1.
In this exercise, assuming the equation is true for n = k means writing: \(\sum_{i=1}^{k} h_{i} = (k+1) h_{k} - k\). The challenge then is to prove it for n = k + 1, which involves showing: \(\sum_{i=1}^{k+1} h_{i} = ((k+1)+1) h_{k+1} - (k+1).\)
Using the sum expansion, \(\sum_{i=1}^{k+1} h_{i} = \sum_{i=1}^{k} h_{i} + h_{k+1}\), you substitute your earlier assumption and manipulate the equation until both sides are shown to be equal. This step rigorously establishes the pattern recognition and logical flow inherent to mathematical induction, enabling a strong grip on the process.