The value of n in the expression `n^2 - 2(n!) + n = 0` for every `n in N` is :

Options

A. a.6
B. b.1
C. c.3
D. d.data insufficient.

Answer

To solve the equation \( n^2 - 2(n!) + n = 0 \) for every \( n \in \mathbb{N} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ n^2 - 2(n!) + n = 0 \]
This can be rearranged to:
\[ n^2 + n - 2(n!) = 0 \]

### Step 2: Isolate the factorial term
We can isolate the factorial term:
\[ 2(n!) = n^2 + n \]
Dividing both sides by 2 gives:
\[ n! = \frac{n^2 + n}{2} \]

### Step 3: Test natural numbers
Now, we will test natural numbers \( n \) to find values that satisfy this equation.

#### Testing \( n = 1 \):
\[ 1! = \frac{1^2 + 1}{2} \implies 1 = \frac{1 + 1}{2} \implies 1 = 1 \quad \text{(True)} \]

#### Testing \( n = 2 \):
\[ 2! = \frac{2^2 + 2}{2} \implies 2 = \frac{4 + 2}{2} \implies 2 = 3 \quad \text{(False)} \]

#### Testing \( n = 3 \):
\[ 3! = \frac{3^2 + 3}{2} \implies 6 = \frac{9 + 3}{2} \implies 6 = 6 \quad \text{(True)} \]

#### Testing \( n = 4 \):
\[ 4! = \frac{4^2 + 4}{2} \implies 24 = \frac{16 + 4}{2} \implies 24 = 10 \quad \text{(False)} \]

#### Testing \( n = 5 \):
\[ 5! = \frac{5^2 + 5}{2} \implies 120 = \frac{25 + 5}{2} \implies 120 = 15 \quad \text{(False)} \]

#### Testing \( n = 6 \):
\[ 6! = \frac{6^2 + 6}{2} \implies 720 = \frac{36 + 6}{2} \implies 720 = 21 \quad \text{(False)} \]

### Step 4: Conclusion
From our testing, the values of \( n \) that satisfy the equation are \( n = 1 \) and \( n = 3 \).

### Final Answer
The values of \( n \) in the expression \( n^2 - 2(n!) + n = 0 \) for every \( n \in \mathbb{N} \) are:
\[ n = 1 \quad \text{and} \quad n = 3 \]