What is the sum of the first 40 positive odd integers?

\(\sum \limits_{k=1}^{40}~2k-1 = \\ \left(2\sum \limits_{k=1}^{40}~k\right) - 40 = \\ 40\cdot 41-40 = 40^2 = 1600\)

The sum of the first N odd numbers is simply N^2 =40^2 =1,600