91Ó°ÊÓ

To find the first \(n\) odd integers, we can write them in a sequence. Odd integers are numbers with the form \(2k-1\), where \(k\) starts from 1. So we can write the first \(n\) odd integers as: \(1, 3, 5, 7, \ldots, (2n-1)\).

Let \(S_n\) be the sum of the first \(n\) odd integers, then we can write \(S_n\) as: \(S_n = 1 + 3 + 5 + 7 + \cdots + (2n-1)\).

To prove that \(S_n = n^2\), we can use the mathematical induction method. 1. Base case (n=1): When \(n=1\), there is only one odd integer, which is 1. Therefore, \(S_1 = 1\), and \(1^2 = 1\). So the equation holds true for \(n=1\). 2. Inductive step: Assume that for some arbitrary value \(k > 1\), \(S_k = k^2\). Now we need to show that for \(k+1\), \(S_{k+1} = (k+1)^2\). First, let's calculate the sum of the first k+1 odd integers, which means we will add the next odd integer \((2k+1)\) to the sum of the first k odd integers, \(S_k\). So, \(S_{k+1} = S_k+(2k+1)\). Since we know that \(S_k = k^2\), we can substitute it into the equation: \(S_{k+1} = k^2 + (2k+1)\). Now we simplify this equation further: \(S_{k+1} = k^2 + 2k + 1\). Now, let's look at the expression for \((k+1)^2\): \((k+1)^2 = k^2 + 2k + 1\). Since both expressions for \(S_{k+1}\) and \((k+1)^2\) are equal, the property is true for the \(k+1\) value, and therefore it is true for all integer values of \(n\). By the mathematical induction, we have proved that the sum of the first \(n\) odd integers is equal to \(n^{2}\).