Problem 9 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-26: Prove that the statement is true for every positive integer \(n\). $$ 1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6} $$

Step-by-Step Solution

Answer

The formula is true for all positive integers \( n \) by induction.

Step 1: Base Case Verification

We start with the base case by verifying the statement for the smallest positive integer, which is \( n = 1 \). The left-hand side of the equation is \( 1^2 = 1 \). The right-hand side of the equation becomes \( \frac{1(1+1)(2\cdot1+1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1 \). Since both sides are equal, the base case holds true.

Step 2: Inductive Hypothesis

Assume that the statement is true for some positive integer \( k \), i.e.,\[ 1^2 + 2^2 + 3^2 + \cdots + k^2 = \frac{k(k+1)(2k+1)}{6}. \]This assumption is called the inductive hypothesis.

Step 3: Inductive Step

We need to prove that if the statement holds for \( n = k \), it also holds for \( n = k+1 \). Consider the sum:\[ 1^2 + 2^2 + 3^2 + \cdots + k^2 + (k+1)^2. \]Using the inductive hypothesis, this can be written as:\[ \frac{k(k+1)(2k+1)}{6} + (k+1)^2. \]

Step 4: Simplifying the Inductive Step

Simplify the expression:\[ \frac{k(k+1)(2k+1)}{6} + (k+1)^2 = \frac{k(k+1)(2k+1) + 6(k+1)^2}{6}. \]Factor out \(k+1\): \[ \frac{(k+1)(k(2k+1) + 6(k+1))}{6}. \]Simplifying the expression inside the parentheses gives:\[ k(2k+1) + 6k + 6 = 2k^2 + k + 6k + 6 = 2k^2 + 7k + 6. \]

Step 5: Final Simplification

Further simplify the factorized form:\[ \frac{(k+1)(2k^2 + 7k + 6)}{6}. \]This can be restructured to:\[ \frac{(k+1)((k+2)(2k+1))}{6}. \]Thus, we have shown that:\[ 1^2 + 2^2 + 3^2 + \cdots + (k+1)^2 = \frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}. \]This matches the original formula for \( n = k+1 \), completing the inductive step.

Step 6: Conclusion

Since the base case is true, and the inductive step has been verified, the formula is true for every positive integer \( n \) by the principle of mathematical induction.

Key Concepts

Base Case

The base case is where we begin when using mathematical induction. It's like checking the first domino in a line to ensure it will fall. We verify that the statement holds true for the smallest value, which is usually 1.
For the given problem, the base case is the first positive integer, which is \( n = 1 \). We calculate both sides of the equation for this value:

• Left-hand side: \( 1^2 = 1 \).
• Right-hand side: \( \frac{1(1+1)(2 \cdot 1+1)}{6} = \frac{1 \cdot 2 \cdot 3}{6} = 1 \).

Both sides are equal, confirming the base case is correct. If this first step holds, it's safe to move forward with the induction process.

Inductive Hypothesis

The inductive hypothesis is our assumption, like guessing that a pattern holds true up to some point. For mathematical induction, it's crucial.
In this case, we assume that the sum formula works for the positive integer \( k \). This means believing the equation:

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

This hypothesis is a stepping stone to proving its truth for the next integer, setting the stage for the inductive step.

Inductive Step

The inductive step is where we prove that if the statement holds for \( n = k \), it also holds for \( n = k+1 \). It’s like causing the next domino to fall after the first one.
To perform the inductive step, we add \( (k+1)^2 \) to both sides of our hypothesis:

• Starting point: \( 1^2 + 2^2 + 3^2 + \cdots + k^2 + (k+1)^2 \).
• Using the hypothesis: \( \frac{k(k+1)(2k+1)}{6} + (k+1)^2 \).

We simplify to express this as a single formula showing the pattern holds for \( k+1 \), ensuring the entire formula continues to work as \( n \) grows.

Sum of Squares Formula

The sum of squares formula is a specific mathematical expression that helps calculate the squares of numbers efficiently.
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 is used to find the sum of squares up to any positive integer \( n \) without adding each square individually.
The formula comes into play during the mathematical induction process, aiding in proving that the pattern remains consistent as \( n \) increases.
Understanding the sum of squares formula helps in recognizing patterns and solving complex mathematical problems more conveniently.