Today we will prove a beautiful mathematical identity using induction. For any positive integer n, the square of the sum of the first n natural numbers equals the sum of their cubes. Let's see an example with n equals 3: the left side gives us 6 squared which is 36, and the right side gives us 1 cubed plus 2 cubed plus 3 cubed, which is 1 plus 8 plus 27, also equal to 36.
Now let's prove this identity using mathematical induction. Induction has two main steps: first, the base case where we prove the statement for n equals 1, and second, the inductive step where we show that if the statement is true for some k, then it must also be true for k plus 1. Let's start with the base case. For n equals 1, the left side is 1 squared which equals 1, and the right side is 1 cubed which also equals 1. Since both sides equal 1, the base case is verified.
Now we move to the inductive step. We assume that our statement P of k is true for some positive integer k. This means we assume that the square of the sum of the first k natural numbers equals the sum of their cubes. This assumption is called the inductive hypothesis. Our goal is to use this assumption to prove that P of k plus 1 is also true. That is, we need to show that the square of the sum of the first k plus 1 natural numbers equals the sum of their cubes including k plus 1 cubed.
Now let's prove the inductive step by showing both sides of P of k plus 1 are equal. We'll start with the left side. The left side is the square of 1 plus 2 plus up to k plus k plus 1. Using the formula for the sum of first n natural numbers, we can write this as k times k plus 1 over 2, plus k plus 1, all squared. Factoring out k plus 1, we get k plus 1 times k over 2 plus 1, squared. Simplifying the inner expression gives us k plus 1 times k plus 2 over 2, all squared. This equals k plus 1 times k plus 2 over 2, quantity squared.
Now for the right side of P of k plus 1. Using our inductive hypothesis, we can replace the sum of cubes from 1 to k with the square of k times k plus 1 over 2. This gives us k squared times k plus 1 squared over 4, plus k plus 1 cubed. Factoring out k plus 1 squared, we get k plus 1 squared times k squared over 4 plus k plus 1. Simplifying the expression in parentheses gives us k squared plus 4k plus 4 over 4, which equals k plus 2 squared over 4. Therefore, the right side also equals k plus 1 times k plus 2 over 2, quantity squared. Since both sides are equal, we have completed the inductive step. By mathematical induction, the identity is proven for all positive integers n.