The given series \( C = 1 + 3 + 5 + 7 + \ldots + 99 \) is an arithmetic series where the first term \( a = 1 \) and the common difference \( d = 2 \).

To find the sum of this series, we can use the formula for the sum of the first \( n \) terms of an arithmetic series:

\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]

First, we need to determine the number of terms \( n \) in the series. The \( n \)-th term of an arithmetic series can be found using the formula:

\[ a_n = a + (n-1)d \]

Given that the last term \( a_n = 99 \):

\[ 99 = 1 + (n-1) \times 2 \]

Solving for \( n \):

\[ 99 = 1 + 2n - 2 \]
\[ 99 = 2n - 1 \]
\[ 100 = 2n \]
\[ n = 50 \]

So, there are 50 terms in the series.

Now, we can find the sum \( S_{50} \):

\[ S_{50} = \frac{50}{2} \times (2 \times 1 + (50-1) \times 2) \]
\[ S_{50} = 25 \times (2 + 49 \times 2) \]
\[ S_{50} = 25 \times (2 + 98) \]
\[ S_{50} = 25 \times 100 \]
\[ S_{50} = 2500 \]

Therefore, the sum of the series \( 1 + 3 + 5 + 7 + \ldots + 99 \) is \( 2500 \).