Proof

Let \(P(n)\) be the statement \(1+3 +5 + \cdots + (2n-1) = n^2\text{.}\) We will prove that \(P(n)\) is true for all \(n \ge 1\text{.}\) First the base case, \(P(1)\text{.}\) We have \(1 = 1^2\) which is true, so \(P(1)\) is established. Now the inductive case. Assume that \(P(k)\) is true for some fixed arbitrary \(k \ge 1\text{.}\) That is, \(1 + 3 + 5 + \cdots + (2k-1) = k^2\text{.}\) We will now prove that \(P(k+1)\) is also true (i.e., that \(1 + 3 + 5 + \cdots + (2k+1) = (k+1)^2\)). We start with the left-hand side of \(P(k+1)\) and work to the right-hand side:

Thus \(P(k+1)\) holds, so by the principle of mathematical induction, \(P(n)\) is true for all \(n \ge 1\text{.}\)