Induction the sum of the first n odd numbers

Statement

Prove by induction that for every integer $n \ge 1$, $1 + 3 + 5 + \cdots + (2n-1) = n^2.$

Solution

Base ($n=1$): the left side is $1$ and the right side is $1^2 = 1$. True.

Inductive step: assume the equality holds for $n = k$, i.e. $1 + 3 + \cdots + (2k-1) = k^2.$ Adding the $(k+1)$-th odd number, which is $2(k+1)-1 = 2k+1$: $1 + 3 + \cdots + (2k-1) + (2k+1) = k^2 + (2k+1) = k^2 + 2k + 1 = (k+1)^2.$ This is the claim for $n = k+1$. By the principle of induction the equality holds for all $n \ge 1$. $\blacksquare$