\[\begin {align} \sum_{k = 1}^n (2k - 1) & = 2\sum_{k = 1}^n k - \sum_{k = 1}^n 1 \\ & = 2 \frac {n(n + 1)} 2 - n \\ & = n(n + 1) - n \\ & = n^2 + n - n \\ & = n^2 \end {align}\]Find a simple formula for \(\sum_{k = 1}^n (2k - 1)\).
\[\begin {align} \sum_{k = 1}^n (2k - 1) & = 2\sum_{k = 1}^n k - \sum_{k = 1}^n 1 \\ & = 2 \frac {n(n + 1)} 2 - n \\ & = n(n + 1) - n \\ & = n^2 + n - n \\ & = n^2 \end {align}\]Find a simple formula for \(\sum_{k = 1}^n (2k - 1)\).