Problem 33 - Calculus: Early Transcendental Functions

Question

Use mathematical induction to prove that \(\sum_{i=1}^{n} i^{3}=\frac{n^{2}(n+1)^{2}}{4}\) for all integers \(n \geq 1 .\)

Step-by-Step Solution

Answer

The principle of mathematical induction is used to prove the given equation. For a base case when \(n = 1\), we substitute \(n = 1\) into the given equation and verify that the equation holds. Then, for the inductive step, we assume that the equation is true for \(n = k\) and then prove it for \(n = k+1\). After simplifying both sides of the equation, we can conclude that if the statement is true for \(n = k\), then it holds true for \(n = k+1\). Hence, the proposition is established for all \(n \geq 1\).

Step 1: Base Case

The first step is to verify the base case (i.e., when \(n = 1\)). Replace \(n\) by \(1\) in the left-hand side (LHS) and right-hand side (RHS) of given equation.LHS: \(\sum_{i=1}^{1} i^{3} = 1^{3} = 1\)RHS: \(\frac{1^{2}(1+1)^{2}}{4} = 1\)Hence, we can see that LHS = RHS, which means our base case holds true.

Step 2: Inductive Step

In the inductive step, we need to prove that if the given equation is true for \(n=k\), it is also true for \(n = k+1\). So let's assume equation is true for \(n = k\). That is,\(\sum_{i=1}^{k} i^{3} = \frac{k^{2}(k+1)^{2}}{4}..... (1)\)We need to prove:\(\sum_{i=1}^{k+1} i^{3} = \frac{(k+1)^{2}(k+2)^{2}}{4}..... (2)\)To achieve this, add \((k+1)^3\) to both sides of equation (1):LHS: \(\sum_{i=1}^{k+1} i^{3} = \sum_{i=1}^{k} i^{3} + (k+1)^3 = \frac{k^{2}(k+1)^{2}}{4} + (k+1)^3 \)Now simplify the RHS of equation (2) and compare it with this resultant LHS. If they match, the induction hypothesis is proved.

Step 3: Simplification and Result

First, let's simplify the RHS of equation (2):RHS: \(\frac{(k+1)^{2}(k+2)^{2}}{4}\)Upon simplification, this becomes \(\frac{k^{2}(k+1)^{2}}{4} + (k+1)^3\)Comparing this with the simplified LHS, we find both match. Hence, we have proved that if the statement is true for \(n = k\), then it is also true for \(n = k+1\).

Key Concepts

Summation

Summation is a mathematical concept used to add up a sequence of numbers. In mathematical notation, summation is represented by the symbol \(\Sigma\). For example, considers the sum \(\sum_{i=1}^{n} i^{3}\), which represents the sum of the cubes of integers from 1 to \(n\). Summation allows us to organize and calculate the collective magnitude of a set of numbers in a concise way.

When dealing with summations, it's important to understand

• The index of summation, which indicates the starting value and the end point.
• The summation expression, which describes how each term in the sequence is calculated.

In our example, \(\sum_{i=1}^{n} i^{3} = i^{3}+ (i+1)^{3}+ \ldots + n^{3}\). This notation simplifies handling potentially lengthy calculations for sequences of numbers.

Cube of Integers

Cubing an integer means raising it to the power of three. For instance, if \(i\) is an integer, then \(i^{3}\) is its cube, calculated as \(i \times i \times i\).

Cubing numbers tends to grow them quickly in size, which is essential when dealing with sequences like \(1^3, 2^3, 3^3, \ldots, n^3\). The sum of these cubes, found using the formula from the exercise, gives us a meaningful total but can be quite large and complex.

This concept is important in various mathematical applications such as calculating volumes and solving complex series, making it a vital mathematical operation for both theoretical and practical purposes.

Inductive Proof

Inductive proof is a technique used in mathematics to prove statements or formulas that claim something is true for all natural numbers. The process involves two major steps:

• First, prove the base case where the statement holds for the initial value of the variable (usually when \(n = 1\)).
• Next, proceed with the inductive step. This involves assuming the statement is true for \(n = k\) and then showing it holds for \(n = k+1\).

The inductive proof method is somewhat like climbing a ladder: get on the first step (base case), then prove that if you can stand on a step, you can move to the next one (inductive step). This shows that if the statement works for one number, it works for the next, and so on for all numbers.

Base Case

The base case is the foundation of the inductive proof process. It involves verifying that a statement is true when \(n\) is at its smallest possible value, often \(n = 1\). This step is crucial because it establishes the starting point from which we can prove the rest of the cases.

For the given exercise, this involves checking if \(\sum_{i=1}^{1} i^{3} = \frac{1^{2}(1+1)^{2}}{4}\). After substituting the value, you see both sides of the equation simplify to 1, confirming the base case.

Without verifying the base case, the inductive proof would lack a reliable starting point, making it impossible to ensure the validity of the statement for all subsequent integers. This makes the base case a cornerstone of any proof by induction.