n is a natural number. It is given that (n +20) + (n +21) + ......+ (n + 100) is a perfect square. Then the least value of n is `"___________"`

Answer

To solve the problem, we need to find the least natural number \( n \) such that the sum \( (n + 20) + (n + 21) + \ldots + (n + 100) \) is a perfect square.

### Step-by-Step Solution:

1. **Identify the number of terms**:
The series starts at \( n + 20 \) and ends at \( n + 100 \). The number of terms can be calculated as follows:
\[ \text{Number of terms} = (100 - 20) + 1 = 81 \]

**Hint**: Count the total number of terms in the series by subtracting the first term from the last term and adding 1.

2. **Calculate the sum of the series**:
The sum \( S \) of the series can be expressed as:
\[ S = (n + 20) + (n + 21) + \ldots + (n + 100) = 81n + (20 + 21 + \ldots + 100) \]

To find \( 20 + 21 + \ldots + 100 \), we use the formula for the sum of an arithmetic series:
\[ \text{Sum} = \frac{\text{Number of terms}}{2} \times (\text{First term} + \text{Last term}) = \frac{81}{2} \times (20 + 100) = \frac{81}{2} \times 120 = 81 \times 60 = 4860 \]
Therefore, the sum becomes:
\[ S = 81n + 4860 \]

**Hint**: Use the formula for the sum of an arithmetic series to simplify the calculation.

3. **Set up the equation for perfect square**:
We need \( S = 81n + 4860 \) to be a perfect square. Let:
\[ S = k^2 \quad \text{for some integer } k \]
Rearranging gives:
\[ 81n + 4860 = k^2 \implies 81n = k^2 - 4860 \implies n = \frac{k^2 - 4860}{81} \]

**Hint**: Rearranging the equation helps to isolate \( n \).

4. **Find conditions for \( n \) to be a natural number**:
For \( n \) to be a natural number, \( k^2 - 4860 \) must be divisible by 81. We can check values of \( k \) such that \( k^2 \) is greater than 4860.

5. **Calculate \( k \)**:
The smallest integer \( k \) such that \( k^2 > 4860 \) can be found by taking the square root:
\[ k > \sqrt{4860} \approx 69.7 \implies k \geq 70 \]
We will check \( k = 70, 71, 72, \ldots \) until we find the smallest \( n \).

6. **Check values of \( k \)**:
- For \( k = 70 \):
\[ k^2 = 4900 \implies n = \frac{4900 - 4860}{81} = \frac{40}{81} \quad \text{(not a natural number)} \]
- For \( k = 71 \):
\[ k^2 = 5041 \implies n = \frac{5041 - 4860}{81} = \frac{181}{81} \quad \text{(not a natural number)} \]
- For \( k = 72 \):
\[ k^2 = 5184 \implies n = \frac{5184 - 4860}{81} = \frac{324}{81} = 4 \quad \text{(natural number)} \]

7. **Conclusion**:
The least value of \( n \) for which the sum is a perfect square is:
\[ \boxed{4} \]