If `n` odd, then `(1+3+5+7+ …..` to `n` terms) is equal to

Options

A. `(n^(2) +1)`
B. `(n^(2)-1)`
C. `n^(2)`
D. `(2n^(2) +1)`

Answer

To solve the problem of finding the sum of the first `n` odd numbers, where `n` is odd, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Sequence**: The sequence of the first `n` odd numbers is:
\[ 1, 3, 5, 7, \ldots \]
This sequence continues until we have `n` terms.

2. **Understand the Formula**: It is a known mathematical result that the sum of the first `n` odd numbers is equal to \( n^2 \). This can be expressed as:
\[ S_n = 1 + 3 + 5 + \ldots + (2n - 1) = n^2 \]
where \( S_n \) is the sum of the first `n` odd numbers.

3. **Apply the Formula**: Since we know that the sum of the first `n` odd numbers is \( n^2 \), we can directly conclude that:
\[ S_n = n^2 \]

4. **Conclusion**: Therefore, if `n` is odd, the sum of the first `n` odd numbers is:
\[ \text{Sum} = n^2 \]

### Final Answer:
The sum of the first `n` odd numbers, where `n` is odd, is equal to \( n^2 \).