For an AP, use the formula \( S_n = \frac{n}{2}(a + l) \) to find the sum.
2300
The sum of the series \(1 + 3 + 5 + \ldots + 99\) is computed as an arithmetic series. The initial term is \(a = 1\) and the common difference is \(d = 2\). The formula for the sum of an arithmetic series is \(S_n = \frac{n}{2} \times (2a + (n-1)d)\).
To find the number of terms \(n\), we use the formula for the \(n\)-th term: \(a_n = a + (n-1)d\). Setting \(a_n = 99\), we get \(99 = 1 + (n-1) \times 2\). Solving for \(n\):
\(98 = 2(n-1)\)
\(49 = n-1\)
\(n = 50\)
Substituting \(n = 50\), \(a = 1\), and \(d = 2\) into the sum formula \(S_n\):
\(S_{50} = \frac{50}{2} \times (2 \times 1 + (50-1) \times 2)\)
\(S_{50} = 25 \times (2 + 98)\)
\(S_{50} = 25 \times 100\)
\(S_{50} = 2500\)
The sum of the series \(1 + 3 + 5 + \ldots + 99\) is \(2500\).