8) Show that the sum of \( n \) terms of the sevies \( 1+3+5+7+\cdots \) is given by Sn \( =n^{2} \) and hence find the sum of the 50 odd numbers \( 1+3,+5+7+\ldots+99 \).

1. The sequence is given by \(1 + 3 + 5 + 7 + \cdots\). Notice that the \(k\)th term of this series is \[ a_k = 2k - 1. \] 2. The sum of the first \(n\) terms, \(S_n\), is \[ S_n = \sum_{k=1}^{n} (2k-1). \] 3. We can split the summation: \[ S_n = 2\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1. \] 4. Recall the formulas: \[ \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \quad \text{and} \quad \sum_{k=1}^{n} 1 = n. \] 5. Substitute these into the expression: \[ S_n = 2\left(\frac{n(n+1)}{2}\right) - n = n(n+1) - n = n^2 + n - n = n^2. \] 6. Thus, the sum of the first \(n\) odd numbers is: \[ S_n = n^2. \] 7. To find the sum of the 50 odd numbers \(1 + 3 + 5 + \cdots + 99\), let \(n = 50\) and substitute into the formula: \[ S_{50} = 50^2 = 2500. \] The sum of the 50 odd numbers is \(2500\).