Question

Prove by induction: (a) \(1^{2}+2^{2}+3^{2}+\cdots+n^{2}=n(n+1)(2 n+1) / 6\) for \(n \geq 0\) (b) \(1^{3}+2^{3}+3^{3}+\cdots+n^{3}=(1+2+3+\cdots+n)^{2}\) for \(n \geq 0\) (c) \(1^{4}+2^{4}+3^{4}+\cdots+n^{4}=n(n+1)(2 n+1)\left(3 n^{2}+3 n-1\right) / 30\) for \(n \geq 0\) (d) \(1^{5}+2^{5}+3^{5}+\cdots+n^{5}=\frac{1}{6} n^{6}+\frac{1}{2} n^{5}+\frac{5}{12} n^{4}-\frac{1}{12} n^{2}\) for \(n \geq 0\)

Short answer

All given formulas are proven true for \(n \geq 0\) by induction.

Step-by-step solution

Base Case for Induction

We first prove the base case for each statement by checking if the given formulas are valid for \(n = 0\). For every equation, when \(n = 0\), both the left-hand side and right-hand side are zero, confirming the base case is true for all statements.

Induction Hypothesis

Assume the formula holds for some arbitrary natural number \(k\). For instance, assume:(a) \(1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}\)(b) \(1^3 + 2^3 + \cdots + k^3 = \left(\frac{k(k+1)}{2}\right)^2\)(c) \(1^4 + 2^4 + \cdots + k^4 = \frac{k(k+1)(2k+1)(3k^2+3k-1)}{30}\)(d) \(1^5 + 2^5 + \cdots + k^5 = \frac{1}{6}k^6 + \frac{1}{2}k^5 + \frac{5}{12}k^4 - \frac{1}{12}k^2\)".

Inductive Step

Prove that if the formula is true for \(n = k\), it must be true for \(n = k+1\). This involves:(a) Adding \((k+1)^2\) to both sides of the assumption and simplifying to match the formula for \(n = k+1\).(b) Adding \((k+1)^3\) and simplifying to show it equals \((1 + 2 + \cdots + (k+1))^2\).(c) Adding \((k+1)^4\) and rearranging to fit the formula pattern for \(n = k+1\).(d) Adding \((k+1)^5\) and proving the expression simplifies to the given polynomial formula for \(n = k+1\).

Step 3a: Verify for \(n = k+1\) for (a)

Add \((k+1)^2\) to the equation in the induction assumption for \(1^2 + 2^2 + \cdots + k^2\), resulting in \(1^2 + 2^2 + \cdots + k^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2\), and show it equals \(\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}\).

Step 3b: Verify for \(n = k+1\) for (b)

Add \((k+1)^3\) to \(1^3 + 2^3 + \cdots + k^3\), simplifying to show it becomes \((1 + 2 + \cdots + (k+1))^2\).

Step 3c: Verify for \(n = k+1\) for (c)

Add \((k+1)^4\) to the existing sum of fourth powers up to \(k\), simplifying to show that it equals \(\frac{(k+1)((k+1)+1)(2(k+1)+1)(3(k+1)^2+3(k+1)-1)}{30}\).

Step 3d: Verify for \(n = k+1\) for (d)

Add \((k+1)^5\) to the sum of fifth powers up to \(k\), and simplify to confirm that it matches \(\frac{1}{6}(k+1)^6 + \frac{1}{2}(k+1)^5 + \frac{5}{12}(k+1)^4 - \frac{1}{12}(k+1)^2\).

Conclusion: Induction Complete

Since all expressions have held for the base case and satisfied the inductive step for \(n = k+1\), the proofs by induction are complete for all given statements (a-d).

Key concepts

Base Case

In mathematical induction, the base case is the foundation. It is the initial step where we verify the formula is true for the smallest natural number, typically 0 or 1. This step is crucial because it establishes that the statement holds true at the starting point of the sequence.

• For each formula given in the exercise, we checked whether the formula holds true for \( n = 0 \).
• In this scenario, when \( n = 0 \), both the left-hand and right-hand sides of the equations are equal to zero.
• This confirms that the base case for all given expressions is valid.

Without affirming the base case, the entire inductive proof might not be trusted since there would be no starting point of validity.

Inductive Step

Once the base case is established, we proceed to the inductive step, which is about proving the 'domino effect'. We assume the formula is true for some arbitrary natural number \( k \). This is called the induction hypothesis.

• The main goal here is to demonstrate that if the formula holds for \( n = k \), then it must also hold for \( n = k+1 \).
• This requires us to add the next term of the series on both sides of the equation and prove the resulting expression aligns with the formula for \( n = k+1 \).

By successfully proving this step, we effectively show that the formula is valid for all natural numbers starting from the base case, creating a chain of validity.

Sum of Powers

The concept of sum of powers involves finding the sum of integers raised to a specific power. This is expressed as \( 1^p + 2^p + 3^p + \cdots + n^p \) for some positive integer \( p \).

• For example, from the exercise: sum of squares, cubes, and higher powers for integers up to \( n \).
• Each of these has a particular closed-form formula, providing us a direct way to compute the sum without adding each term separately.

Understanding these formulas is important for simplifying complex problems and finding patterns in algebra and calculus.

Natural Numbers

Natural numbers are the set of positive integers beginning from 0 or 1 and extending indefinitely. They are denoted usually by \( \mathbb{N} \). Here's why they are essential:

• They form the building blocks for arithmetic and number theory.
• In induction, the base case and the inductive step usually rely on these numbers as they provide a framework to proceed from the simplest form to a generalized statement.
• The formulas being proven by induction in this context are specifically applicable to natural numbers.

Thus, the sequence from 0 or 1 plays a critical role in validating mathematical propositions through induction.