Question

Write out the information that describes what the inductive step assumes and what the step must prove for proving $$1^{5}+2^{5}+3^{5}+\cdots+n^{5}=\frac{1}{6} n^{6}+\frac{1}{2} n^{5}+\frac{5}{12} n^{4}-\frac{1}{12} n^{2}$$ with \(n_{0}\) given.

Short answer

Assume true for \( n = k \) and prove for \( n = k+1 \).

Step-by-step solution

Base Case Preparation

To begin the induction, ensure you know the base value, usually \( n = 1 \), and evaluate the expression for this value to see if the equality holds. Check if \( 1^5 = \frac{1}{6} \times 1^6 + \frac{1}{2} \times 1^5 + \frac{5}{12} \times 1^4 - \frac{1}{12} \times 1^2 \). Calculate both sides to confirm they match.

Inductive Hypothesis Explanation

Assume the statement is true for some arbitrary natural number \( n = k \). That is, assume:\[1^{5} + 2^{5} + \.\.\. + k^{5} = \frac{1}{6} k^{6} + \frac{1}{2} k^{5} + \frac{5}{12} k^{4} - \frac{1}{12} k^{2}\]This assumption is called the inductive hypothesis.

Inductive Step Setup

For the inductive step, we must prove that the statement is also true for \( n = k + 1 \). Start by writing what needs to be proved:\[1^{5} + 2^{5} + \.\.\. + k^{5} + (k+1)^{5} = \frac{1}{6} (k+1)^{6} + \frac{1}{2} (k+1)^{5} + \frac{5}{12} (k+1)^{4} - \frac{1}{12} (k+1)^{2}\]

Substitute the Hypothesis

Using the inductive hypothesis, substitute for the left-hand side up to \( k \):\[\frac{1}{6} k^{6} + \frac{1}{2} k^{5} + \frac{5}{12} k^{4} - \frac{1}{12} k^{2} + (k+1)^{5}\]This should equal the expression for \( n = k+1 \).

Simplify and Prove Equality

Next, compute \((k+1)^5\) and input it into the expanded expression. Keep simplifying the entire expression until you show that it rearranges identically to resemble \(\frac{1}{6} (k+1)^{6} + \frac{1}{2} (k+1)^{5} + \frac{5}{12} (k+1)^{4} - \frac{1}{12} (k+1)^{2}\). This proves the statement for \( n = k + 1 \).

Conclusion of Proof

Since the base case \( n = 1 \) is verified and the inductive step is proven, by the principle of mathematical induction, the formula holds for all natural numbers \( n \geq 1 \).

Key concepts

Base Case

The base case is where any inductive proof begins. It's like laying the foundation of a house. Without a strong foundation, the house cannot stand, just as the rest of the induction wouldn't hold if the base case fails. In mathematical induction, the base case often starts with the smallest natural number, typically when \( n = 1 \).

To illustrate this with our specific problem, we need to confirm the equality:

• Left side: \( 1^5 \)
• Right side: \( \frac{1}{6} \times 1^6 + \frac{1}{2} \times 1^5 + \frac{5}{12} \times 1^4 - \frac{1}{12} \times 1^2 \)

After computing both sides, they should be equal, validating that the base case holds true. Without completing this check, the whole inductive proof would be built on an uncertain premise.

Inductive Hypothesis

The inductive hypothesis is a critical step, as it sets the stage for proving the statement for all natural numbers. It requires us to assume the formula holds true for some particular value \( n = k \). This assumption isn't random but a deliberate wager that allows us to bridge to proving the case for \( n = k+1 \).

In our example, we assume:

• \( 1^{5} + 2^{5} + \ldots + k^{5} = \frac{1}{6} k^{6} + \frac{1}{2} k^{5} + \frac{5}{12} k^{4} - \frac{1}{12} k^{2} \)

This step does not require proof but is pivotal as it forms the hypothesis upon which the next step, the inductive step, relies.

Inductive Step

The inductive step is where the magic of mathematical induction happens. Here, we use the inductive hypothesis to establish that if the statement works for \( n = k \), then it must also work for \( n = k+1 \). This logical leap relies on our prior hypothesis being correct to project correctness onto the subsequent number.

For our specific problem, the inductive step requires proving:

• \( 1^{5} + 2^{5} + \ldots + k^{5} + (k+1)^{5} = \frac{1}{6} (k+1)^{6} + \frac{1}{2} (k+1)^{5} + \frac{5}{12} (k+1)^{4} - \frac{1}{12} (k+1)^{2} \)

By substituting the inductive hypothesis into the left side of the equation and simplifying, we aim to show that both sides are indeed equal, proving the step.

Polynomial Equations

Polynomial equations play a crucial role in many mathematical proofs, including the one in our discussion. A polynomial equation involves terms of variables raised to whole number powers. The degree of the polynomial is the highest power of the variable present in the expression.

In the exercise, we encounter polynomials like:

• \( \frac{1}{6} n^{6} \)
• \( \frac{1}{2} n^{5} \)
• \( \frac{5}{12} n^{4} \)
• \( -\frac{1}{12} n^{2} \)

Each term is a piece of this polynomial puzzle. When solving proofs involving polynomials, understanding their behavior as \( n \) changes is critical. The expansion of polynomial terms, especially during the inductive step, is essential to demonstrate the validity of the statement for \( n = k+1 \). Understanding polynomials allows us to confidently manipulate and equate sides of equations in these proofs.