Proving the Sum of Squares Using Mathematical Induction
In this article, we will explore the formula for the sum of the squares of the first n natural numbers using mathematical induction. The formula we are aiming to prove is:
Formula and Notations
The formula for the sum of the squares of the first n natural numbers is:
(1^2 2^2 3^2 ldots n^2 frac{n(n 1)(2n 1)}{6})
Mathematical Induction
Mathematical induction is a powerful method for proving propositions that are indexed by natural numbers. The process involves two main steps: the base case and the inductive step. Here, we will prove the given formula using these steps.
Step 1: Base Case
First, we check the base case for n 1: [1^2 1]
The formula gives:
[1^2 2^2 3^2 ldots 1^2 frac{1(1 1)(2 cdot 1 1)}{6} frac{1 cdot 2 cdot 3}{6} 1]Since both sides are equal, the base case holds.
Step 2: Inductive Step
Assume the formula holds for some integer k that is:
[1^2 2^2 3^2 ldots k^2 frac{k(k 1)(2k 1)}{6}]We need to show that it also holds for k 1:
[1^2 2^2 3^2 ldots k^2 (k 1)^2 frac{(k 1)(k 2)(2(k 1) 1)}{6}]Step 3: Use the Inductive Hypothesis
Starting from the left side of the equation for k 1:
[1^2 2^2 3^2 ldots k^2 (k 1)^2 frac{k(k 1)(2k 1)}{6} (k 1)^2]Now we can factor out k 1:
[ frac{k(k 1)(2k 1)}{6} frac{6(k 1)^2}{6}]Combining the fractions:
[ frac{k(k 1)(2k 1) 6(k 1)^2}{6}]Step 4: Simplify the Expression
Now simplify the expression inside the parenthesis:
[k(k 1)(2k 1) 6(k 1)^2 2k^2 k 6k^2 12k 6]8k^2 13k 6]
Step 5: Substitute Back
Now substitute this back into our equation:
[ frac{(k 1)(8k^2 13k 6)}{6}]Step 6: Final Form
This matches the right-hand side we wanted to prove:
[frac{(k 1)((k 1) 1)(2(k 1) 1)}{6} frac{(k 1)(k 2)(2k 3)}{6}]Thus, we have shown that if the formula holds for n k it also holds for n k 1.
Conclusion
By the principle of mathematical induction, the formula (1^2 2^2 3^2 ldots n^2 frac{n(n 1)(2n 1)}{6}) is true for all positive integers n.