The solution of the equation `1!+2!+3!+…+x!=k^(2)` where `k in l` are

Options

A. 1, 3
B. 2, 3
C. 4, 5
D. `-1, -2`

Answer

To solve the equation \(1! + 2! + 3! + \ldots + x! = k^2\) where \(k\) is an integer, we will use a trial and error method to find the values of \(x\) for which the left-hand side is a perfect square.

### Step-by-Step Solution:

1. **Start with \(x = 1\)**:
\[ 1! = 1 \]
Here, \(k^2 = 1\), which gives \(k = \pm 1\). Since \(k\) is an integer, \(x = 1\) is a valid solution.

**Hint**: Check if the result is a perfect square.

2. **Next, try \(x = 2\)**:
\[ 1! + 2! = 1 + 2 = 3 \]
Here, \(k^2 = 3\), which gives \(k = \pm \sqrt{3}\). Since \(\sqrt{3}\) is not an integer, \(x = 2\) is not a valid solution.

**Hint**: Check if \(k^2\) results in a perfect square.

3. **Now, try \(x = 3\)**:
\[ 1! + 2! + 3! = 1 + 2 + 6 = 9 \]
Here, \(k^2 = 9\), which gives \(k = \pm 3\). Since \(k\) is an integer, \(x = 3\) is a valid solution.

**Hint**: Continue testing higher values of \(x\).

4. **Next, try \(x = 4\)**:
\[ 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 \]
Here, \(k^2 = 33\), which gives \(k = \pm \sqrt{33}\). Since \(\sqrt{33}\) is not an integer, \(x = 4\) is not a valid solution.

**Hint**: Verify if the sum is a perfect square.

5. **Now, try \(x = 5\)**:
\[ 1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153 \]
Here, \(k^2 = 153\), which gives \(k = \pm \sqrt{153}\). Since \(\sqrt{153}\) is not an integer, \(x = 5\) is not a valid solution.

**Hint**: Keep checking for higher values of \(x\) if needed.

6. **Conclusion**:
The valid solutions for \(x\) where \(1! + 2! + 3! + \ldots + x! = k^2\) are \(x = 1\) and \(x = 3\).

### Final Answer:
The values of \(x\) for which the equation holds true are \(1\) and \(3\).