In this case, we have `a_1 = 1`, `d = 2` and `n = 1000.`

So, using the formula

`S_n=n/2[2a_1+(n-1)d]`

the sum `1 + 3 + 5 + ...` for 1000 steps is given by:

`S_1000` `=1000/2[2(1)+(1000-1)(2)]` `=1\ 000\ 000`