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 $$ 1+3,+5+7+\ldots+99 $$.

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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 $$2500$$.
The sum of the first $$n$$ odd numbers is $$n^2$$. Therefore, the sum of the first 50 odd numbers is $$50^2 = 2500$$.

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 \( 1+3,+5+7+\ldots+99 \).

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