Proving the Sum of Squares Formula Using Mathematical Induction

The formula we aim to prove is the well-known identity that the sum of the squares of the first n natural numbers can be expressed as:

(1^2 2^2 3^2 ldots n^2 frac{n(n 1)(2n 1)}{6})

Introduction to the Problem

This article will focus on the proof of the above identity using mathematical induction. Mathematical induction is a powerful tool in number theory and algebra that allows us to prove statements for an infinite sequence of natural numbers. It relies on the following two steps:

The base case: We prove the statement for the smallest value of (n). The inductive step: We assume the statement is true for some (k) and prove it is true for (k 1).

Mathematical Induction Base Case

First, let's establish the base case where (n 1).

Base Case (n1):LHS  1^2  1RHS  frac{1(1 1)(2(1) 1)}{6}  frac{1 cdot 2 cdot 3}{6}  1LHS  RHS

The base case holds true, as both the left-hand side (LHS) and right-hand side (RHS) yield the value 1.

Induction Hypothesis

Assume the statement is true for (n k), i.e., (1^2 2^2 3^2 ldots k^2 frac{k(k 1)(2k 1)}{6})

Induction Step

To show the statement is true for (n k 1), we need to prove:

(1^2 2^2 3^2 ldots k^2 (k 1)^2 frac{(k 1)(k 2)(2(k 1) 1)}{6})

We start from the induction hypothesis and add ((k 1)^2) to both sides:

(1^2 2^2 3^2 ldots k^2 (k 1)^2 frac{k(k 1)(2k 1)}{6} (k 1)^2)

Factor out ((k 1)) on the right-hand side:

(1^2 2^2 3^2 ldots k^2 (k 1)^2 frac{k(k 1)(2k 1) 6(k 1)^2}{6})

Simplify the expression inside the parentheses:

(frac{k(k 1)(2k 1) 6(k 1)^2}{6} frac{(k 1)[k(2k 1) 6(k 1)]}{6})

Expand and combine like terms:

(frac{(k 1)[2k^2 k 6k 6]}{6} frac{(k 1)(2k^2 7k 6)}{6})

Note that (2k^2 7k 6 (k 1)(2k 6) (k 1)(2k 3)(k 2) / 2), so we simplify:

(frac{(k 1)(k 2)(2k 3)}{6})

Which is exactly the formula for (n k 1).

Hence, the induction step holds true.

Conclusion

We have shown that the sum of the squares of the first (n) natural numbers can be expressed as:

(1^2 2^2 3^2 ldots n^2 frac{n(n 1)(2n 1)}{6})

This formula is correct and has been proven using mathematical induction.

Reference:

1. Square Pyramidal Number