Find a simple formula for \(\sum_{k=1}^{n}(2k-1)\).

Solved by using the arithmetic series identity: \(\sum\limits_{k=1}^{n}k = \frac{1}{2}n(n + 1)\).

\[\begin{split} \sum\limits_{k=1}^{n}(2k-1) & = 2\sum\limits_{k=1}^{n}k - \sum\limits_{k=1}^{n} 1 \\ & = 2 \cdot \frac{1}{2}n(n+1) - n \\ & = n^2 + n - n \\ & = n^2 \\ \end{split}\]