Given:

• First term \( \rm a_1 = -3 \)
• Common difference \( \rm d = 1 - (-3) = 4 \)

We can use the formula for the sum of the first \( \rm n \) terms of an arithmetic series:

\( \rm S_n = \frac{n}{2}(2a_1 + (n - 1)d) \)

Substitute the given values:

\( \rm S_{50} = \frac{50}{2}(2 \times (-3) + (50 - 1) \times 4) \)

\( \rm S_{50} = 25(-6 + 49 \times 4) \)

\( \rm S_{50} = 25(-6 + 196) \)

\( \rm S_{50} = 25 \times 190 \)

\( \rm S_{50} = 4750 \)

So, the sum of the series \(-3 + 1 + 5 + \ldots\) up to 50 terms is \(4750\).