(related to Proposition: Sum of Consecutive Odd Numbers)
We provide a proof by induction for all natural numbers $n.$ * Base case: $n=1$ $$\sum_{k=1}^1(2k-1)=2\cdot 1-1=1=1^2.$$ * Induction step $n\to n+1$ * Assume, $$\sum_{k=1}^n (2k-1)=n^2$$ is correct for an $n\ge 1.$ * Then $$\begin{align}\sum_{k=0}^{n+1} (2k-1)&=n^2+(2(n+1)-1)\nonumber\\ &=n^2+2n+1\nonumber\\ &=(n+1)^2.\nonumber\end{align}$$ * Thus, the sum of consecutive odd numbers from $1$ to $n$ equals the square of $n.$
