What is the square root of the sum of first 36 odd natural numbers?

Options

A. 6
B. 36
C. 64
D. 72

Answer

To find the square root of the sum of the first 36 odd natural numbers, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the first 36 odd natural numbers**:
The first 36 odd natural numbers are: 1, 3, 5, 7, ..., up to the 36th term.

2. **Determine the formula for the nth odd natural number**:
The nth odd natural number can be expressed as:
\[ a_n = 2n - 1 \]
where \( n \) is the term number.

3. **Find the 36th odd natural number**:
Substitute \( n = 36 \) into the formula:
\[ a_{36} = 2(36) - 1 = 72 - 1 = 71 \]

4. **Calculate the sum of the first n odd natural numbers**:
The sum of the first \( n \) odd natural numbers is given by the formula:
\[ S_n = n^2 \]
For \( n = 36 \):
\[ S_{36} = 36^2 = 1296 \]

5. **Find the square root of the sum**:
Now, we need to find the square root of the sum:
\[ \sqrt{S_{36}} = \sqrt{1296} \]
The square root of 1296 is:
\[ \sqrt{1296} = 36 \]

### Final Answer:
The square root of the sum of the first 36 odd natural numbers is **36**.