The expression `1! + 2! + 3! + 4! + ……….. + n!` (where `n ge 5)` is not a/an :

Options

A. composite number
B. odd number
C. perfect number
D. multiple of 3

Answer

To solve the problem of determining what the expression \(1! + 2! + 3! + 4! + \ldots + n!\) (where \(n \geq 5\)) is not, we will follow these steps:

### Step 1: Calculate the factorial values
We start by calculating the factorial values for \(1!\) to \(5!\):
- \(1! = 1\)
- \(2! = 2\)
- \(3! = 6\)
- \(4! = 24\)
- \(5! = 120\)

### Step 2: Sum the factorials
Now we sum these values:
\[ 1! + 2! + 3! + 4! + 5! = 1 + 2 + 6 + 24 + 120 = 153 \]

### Step 3: Analyze the result
Next, we need to analyze the result \(153\) to determine its properties:
- **Odd or Even**: \(153\) is odd.
- **Composite or Prime**: \(153\) is composite because it can be divided by \(3\) (since \(1 + 5 + 3 = 9\) is divisible by \(3\)).
- **Perfect Square**: \(153\) is not a perfect square, as there is no integer \(k\) such that \(k^2 = 153\).
- **Multiple of 3**: Since \(153\) is divisible by \(3\), it is a multiple of \(3\).

### Step 4: Conclusion
From our analysis, we conclude that the expression \(1! + 2! + 3! + 4! + \ldots + n!\) (for \(n \geq 5\)) is not a perfect square number.

### Final Answer
The expression \(1! + 2! + 3! + 4! + \ldots + n!\) (where \(n \geq 5\)) is not a perfect square.

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