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.