The sum of first n odd natural numbers is

Options

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

Answer

To find the sum of the first \( n \) odd natural numbers, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Odd Natural Numbers**:
The first few odd natural numbers are: 1, 3, 5, 7, 9, ..., and so on. The \( n \)-th odd natural number can be expressed as \( 2n - 1 \).

2. **Writing the Sum**:
The sum of the first \( n \) odd natural numbers can be written as:
\[ S = 1 + 3 + 5 + 7 + ... + (2n - 1) \]

3. **Using a Formula**:
There is a well-known formula that states that the sum of the first \( n \) odd natural numbers is equal to \( n^2 \). We can verify this by calculating the sum for small values of \( n \).

4. **Verification**:
- For \( n = 1 \):
\[ S = 1 = 1^2 \]
- For \( n = 2 \):
\[ S = 1 + 3 = 4 = 2^2 \]
- For \( n = 3 \):
\[ S = 1 + 3 + 5 = 9 = 3^2 \]
- For \( n = 4 \):
\[ S = 1 + 3 + 5 + 7 = 16 = 4^2 \]

5. **Conclusion**:
From the above calculations, we can see that the sum of the first \( n \) odd natural numbers is indeed \( n^2 \).

### Final Result:
Thus, the sum of the first \( n \) odd natural numbers is:
\[ S = n^2 \]