Proving the Sum of Squares Using Mathematical Induction

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.