Question

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

Short answer

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

Step-by-step solution

Base Case Verification

The base case is verified when \(n = n_0\). If we assume \(n_0 = 1\), then we need to show that the formula holds for this initial case. The left side of the equation becomes \(1^2 = 1\), and the right side is \((1)(1+1)(2(1)+1)/6 = 1\). Since both sides are equal, the base case holds.

Inductive Hypothesis

Assume that the formula holds true for a particular value \(n = k\). In other words, assume that \(1^2 + 2^2 + 3^2 + \ldots + k^2 = \frac{k(k+1)(2k+1)}{6}\). This assumption is called the inductive hypothesis.

Inductive Step

Based on the inductive hypothesis, we need to show that the formula also holds true for \(n = k + 1\). Start by adding \((k+1)^2\) to the both sides of the inductive hypothesis: \[1^2 + 2^2 + 3^2 + \cdots + k^2 + (k+1)^2 = \frac{k(k+1)(2k+1)}{6} + (k+1)^2\]. Our goal is to show that the right side equals \(\frac{(k+1)(k+2)(2(k+1)+1)}{6}\).

Simplifying the Inductive Step

Calculate \(\frac{k(k+1)(2k+1)}{6} + (k+1)^2\) and try to simplify it to match \(\frac{(k+1)(k+2)(2(k+1)+1)}{6}\). This involves expanding and combining like terms: \[(k+1)^2 = (k+1)(k+1) = k^2 + 2k + 1\], and \[\frac{(k+1)(2k^2 + 7k + 6)}{6}\]. Simplifying confirms the equality and completes the inductive step.

Key concepts

Inductive Hypothesis

The inductive hypothesis is a key concept in mathematical induction.It involves assuming that a given mathematical statement is true for a specific value \( n = k \).This assumption paves the way for proving that the statement holds true for the next integer, \( n = k + 1 \).It is essential because it forms the foundation for the truth of the statement for subsequent values of \( n \).
In this exercise, the inductive hypothesis involves assuming that the sum of squares formula holds true for \( n = k \).This assumption is expressed as:

• \( 1^2 + 2^2 + 3^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6} \)

Meaning, we accept as given that this formula correctly sums up the squares up to \( n = k \).Having this assumption allows us to use it as a steppingstone in the inductive step of the proof.

Base Case

The base case is the first step in mathematical induction.It involves verifying that the given statement is true for the initial value, often starting with the smallest integer, such as \( n_0 = 1 \).Confirming the base case ensures that the formula works at least once, providing a starting point for induction.
In this exercise, the base case checks the formula:

• \( 1^2 = 1 \)

This matches the right side of the equation for \( n_0 = 1 \), which is \( \frac{1 \, (1+1) \, (2\cdot1 + 1)}{6} = 1 \).Since both sides are equal, it confirms that the formula holds at \( n = 1 \).This successful verification is crucial as it forms the foundation of the inductive proof, assuring us that we are starting from a correct point.

Inductive Step

The inductive step is where the power of induction truly shines.It involves proving that if the given statement works for some integer \( n = k \), then it must also work for the next integer \( n = k + 1 \).This step provides the mechanism to extend the validity of the statement from \( k \) to \( k+1 \) and thus to all integers starting from the base case.
To execute this step in our exercise, start with the inductive hypothesis:

• Assume: \( 1^2 + 2^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6} \)
• Add \((k+1)^2\) to both sides: \( 1^2 + 2^2 + \cdots + k^2 + (k+1)^2 \)

This must equal:

• \(\frac{k(k+1)(2k+1)}{6} + (k+1)^2\).

The goal is to manipulate and simplify this equation to arrive at:

• \( \frac{(k+1)(k+2)(2(k+1)+1)}{6} \)

This proves that the statement holds for \( n = k + 1 \), thus completing the inductive step.

Sum of Squares

The sum of squares in mathematics often involves calculating the sum of squared terms, usually starting from 1 up to a specific number \( n \).This particular sequence of squared terms is common in various mathematical problems and can be elegantly expressed using formulas.
In this exercise, the sum of squares is represented as follows:

• \( 1^2 + 2^2 + 3^2 + \cdots + n^2 \)

The formula for this sum of squares is given by:

• \( \frac{n(n+1)(2n+1)}{6} \)

It is a powerful formula because it offers a direct way to calculate the sum of squares without manually adding each squared term.It emerges in proofs like this one to aid in verifying the equality for all integer values using induction.Thus, the sum of squares is not just a calculation, but a steppingstone in forming general statements and proofs in mathematics.