To find the sum of the arithmetic series 1 + 3 + 5 + ... + 99, we can use the formula for the sum of an arithmetic series:

S = (n/2)(a + l)

where S is the sum, n is the number of terms, a is the first term, and l is the last term.

In this case, the first term (a) is 1, the last term (l) is 99, and the common difference (d) is 2 (since each term increases by 2).

We can find the number of terms (n) using the formula:

l = a + (n-1)d

99 = 1 + (n-1)(2)
99 = 1 + 2n - 2
99 = 2n - 1
100 = 2n
n = 50

Now we can substitute the values into the sum formula:

S = (n/2)(a + l)
S = (50/2)(1 + 99)
S = 25(100)
S = 2500

Therefore, the sum of the arithmetic series 1 + 3 + 5 + ... + 99 is 2500.