Problem 10 - A Graphical Approach to College Algebra

Question

Use mathematical induction to prove each statement. Assume that \(n\) is a positive integer. $$1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6}$$

Step-by-Step Solution

Answer

The formula is true by mathematical induction.

Step 1: Base Case

For mathematical induction, we start by proving the base case. Let's test the statement for the smallest positive integer, which is \(n = 1\). Substitute \(n = 1\) into the formula:\[1^2 = \frac{1(1+1)(2 \cdot 1+1)}{6}\]Calculating both sides gives:\[1 = \frac{1 \times 2 \times 3}{6} = \frac{6}{6} = 1\]Since both sides are equal, the base case holds true.

Step 2: Inductive Hypothesis

Assume that the given formula is true for \(n = k\), where \(k\) is any positive integer. This means:\[1^2 + 2^2 + 3^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}\]This hypothesis will be used to prove that the formula also holds for \(n = k+1\).

Step 3: Inductive Step

Now we need to show that if the formula is true for \(n = k\), it must also be true for \(n = k+1\). This translates to: \[1^2 + 2^2 + 3^2 + \cdots + k^2 + (k+1)^2 = \frac{(k+1)(k+2)(2(k+1)+1)}{6}\]Substitute the inductive hypothesis into the above statement:\[\frac{k(k+1)(2k+1)}{6} + (k+1)^2\]

Step 4: Simplifying the Inductive Step Formula

Let's simplify the right-hand side of the equality we want to prove:\[\frac{k(k+1)(2k+1)}{6} + (k+1)^2\]First, rewrite \((k+1)^2\):\[(k+1)^2 = (k+1)(k+1)\]Combine in the formula:\[= (k+1) \left( \frac{k(2k+1)}{6} + (k+1) \right)\]\[= (k+1) \left( \frac{2k^2 + k}{6} + \frac{6(k+1)}{6} \right)\]\[= (k+1) \left( \frac{2k^2 + k + 6k + 6}{6} \right)\]\[= (k+1) \left( \frac{2k^2 + 7k + 6}{6} \right)\]\[= \frac{(k+1)(k+2)(2k+3)}{6}\]This simplifies to exactly the form we wanted to show it equals: \[\frac{(k+1)(k+2)(2(k+1)+1)}{6}\]

Step 5: Conclusion

Since we have proven the base case and shown that if the formula holds for \(n = k\), it also holds for \(n = k+1\), by mathematical induction, the statement\[1^2+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}\] is true for all positive integers \(n\).

Key Concepts

base case

The base case is the starting point of a mathematical induction proof. It involves verifying that a statement is true for the smallest value of the variable involved, often the smallest positive integer.
In this exercise, the base case is checked for when the variable \( n \) equals 1. We substitute \( n = 1 \) into the given formula to see if both sides of the equation match:

• Left side: \( 1^2 = 1 \)
• Right side: \( \frac{1(1+1)(2\cdot1+1)}{6} \)
• Simplifying the right side, we get \( \frac{1 \times 2 \times 3}{6} = 1 \)

Since both the left and the right side equal 1, the base case holds true.
Confirming this step establishes a foundation to start the induction process, ensuring that our statement has at least one correct instance.

inductive hypothesis

The inductive hypothesis involves assuming the given statement is true for an arbitrary but specific positive integer \( k \). This assumption is crucial as it forms the basis to prove the next step in the sequence.
In this problem, we assume the formula holds when \( n = k \). This means we hypothesize that:

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

By accepting this assumption, we prepare to demonstrate that the statement also holds for the next integer, \( k+1 \).
It's crucial at this stage to understand that we're not proving the truth of the statement for \( k \), but assuming it holds to move forward in the induction process.

inductive step

The inductive step is where we extend the assumption of the inductive hypothesis to prove the statement for the next integer, \( n = k + 1 \). It's the heart of mathematical induction, demonstrating that "if it works for one, it will work for the next."
To prove the next case using our hypothesis, we let:

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

We then manipulate this algebraically by combining terms, simplifying factorial expressions, or changing forms to match the structure on the right-hand side.
After careful simplification, confirm both sides are equal, proving that the statement for \( n = k + 1 \) follows from \( n = k \). Once this is established, it demonstrates a chain reaction that completes the induction.

sum of squares

The sum of squares is a mathematical concept referring to the total of each term squared in a sequence from 1 to \( n \). It commonly appears in mathematical problems involving sequences and series.
In this exercise, the formula given is:

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

This formula offers a closed expression for calculating the sum of squares for any positive integer \( n \). The formula's derivation and validity are confirmed using mathematical induction, which involves examining initial cases and using logical steps to extend the truth through all natural numbers.
Understanding and applying this formula can simplify computations and illustrate connections within mathematical structures, making the sum of squares a versatile tool in both theoretical and applied mathematics.