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Chapter 11: Problem 19

The following formulas can be used to find sums of powers of natural numbers. Use mathematical induction to prove each formula. $$1^{3}+2^{3}+3^{3}+\cdots+n^{3}=\frac{n^{2}(n+1)^{2}}{4}$$

Short Answer

Expert verified
Proven by induction.

Step by step solution

01

Base Case

Check the formula for the base case when n = 1. The left-hand side of the equation is: \[1^3 = 1\]The right-hand side of the equation is:\[\frac{1^2(1+1)^2}{4} = \frac{1 \cdot 4}{4} = 1\]Since both sides are equal, the formula holds for n = 1.
02

Inductive Hypothesis

Assume the formula holds for some arbitrary positive integer k. This means: \[1^{3}+2^{3}+3^{3}+\cdots+k^{3}=\frac{k^{2}(k+1)^{2}}{4}\]This is the inductive hypothesis.
03

Inductive Step

We need to show that the formula holds for k + 1. Consider the sum of cubes from 1 to k+1: \[1^3 + 2^3 + 3^3 + \cdots + k^3 + (k+1)^3\]Using the inductive hypothesis, this can be written as: \[\frac{k^{2}(k+1)^{2}}{4} + (k+1)^3\]Simplify the right-hand side: \[\frac{k^{2}(k+1)^{2} + 4(k+1)^3}{4} = \frac{(k+1)^{2}(k^2 + 4(k+1))}{4}\]Combine and simplify inside the parenthesis: \[\frac{(k+1)^{2}(k^2 + 4k + 4)}{4} = \frac{(k+1)^{2}(k + 2)^{2}}{4}\]Hence, \[1^{3} + 2^{3} + 3^{3} + \cdots + k^{3} + (k+1)^{3} = \frac{(k+1)^{2}(k+2)^{2}}{4}\]So the formula holds for k + 1.
04

Conclusion

Since the formula holds for n = 1 (base case) and assuming it holds for an arbitrary k leads to it holding for k + 1, by mathematical induction, the formula is proven true for all natural numbers n.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Sum of Cubes
The sum of cubes is a mathematical concept that sums up the cubes of the first n natural numbers. The formula we will prove by mathematical induction is:
\(1^{3} + 2^{3} + 3^{3} + \text{...} + n^{3} = \frac{n^{2}(n+1)^{2}}{4}\).
Understanding how to prove this formula helps in grasping deeper concepts of algebra and sequences. To prove it, we will use mathematical induction. This proof is broken down into manageable steps to simplify understanding.
Base Case
In any mathematical induction proof, we start with the base case. This is to show that the formula works for the first natural number, which is usually n = 1.

For our formula:
- The left-hand side is \(1^3 = 1\).
- The right-hand side is \(\frac{1^2(1+1)^2}{4} = \frac{1 \times 4}{4} = 1\).
Both sides are equal, proving the formula holds true for n = 1. Getting the base case right is crucial since it forms the foundation of our induction proof.
Inductive Hypothesis
Next, we assume that the formula holds for some arbitrary positive integer, let's call it k. This assumption is known as the inductive hypothesis. Here, we assume:
\(1^{3} + 2^{3} + 3^{3} + \text{...} + k^{3} = \frac{k^{2}(k+1)^{2}}{4}\).
This step doesn't prove anything yet; it sets up the next step. The hypothesis is like taking a logical leap, assuming the formula works for k to then show it must also work for k+1. This will help us in the inductive step.
Inductive Step
Now, we need to prove that if the formula holds for k, it also holds for k+1. We add \((k+1)^3\) to both sides of the inductive hypothesis:
\(1^3 + 2^3 + 3^3 + \text{...} + k^3 + (k+1)^3\).
Using the inductive hypothesis:
\(\frac{k^{2}(k+1)^{2}}{4} + (k+1)^3\).
We simplify the right-hand side:
\(\frac{k^{2}(k+1)^{2} + 4(k+1)^3}{4} = \frac{(k+1)^{2}(k^{2} + 4(k+1))}{4}\), further simplifying to:
\(\frac{(k+1)^{2}(k+2)^{2}}{4}\).
This shows that:
\(1^{3} + 2^{3} + 3^{3} + \text{...} + k^{3} + (k+1)^{3} = \frac{(k+1)^{2}(k+2)^{2}}{4}\).
Therefore, if the formula holds for k, it must hold for k+1. Having proved this, the formula is true by mathematical induction.

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