If `(n+1)! = 6[(n-1)!]`. Find n

Options

A. 6
B. 4
C. 8
D. 2

Answer

To solve the equation \((n+1)! = 6(n-1)!\), we can follow these steps:

### Step 1: Rewrite the factorials
We start by rewriting \((n+1)!\) in terms of \(n\):
\[ (n+1)! = (n+1) \cdot n \cdot (n-1)! \]
So, we can substitute this into our equation:
\[ (n+1) \cdot n \cdot (n-1)! = 6(n-1)! \]

### Step 2: Cancel out \((n-1)!\)
Since \((n-1)!\) appears on both sides of the equation and is not equal to zero (for \(n \geq 2\)), we can divide both sides by \((n-1)!\):
\[ (n+1) \cdot n = 6 \]

### Step 3: Expand and rearrange the equation
Now, we can expand the left-hand side:
\[ n^2 + n = 6 \]
Next, we rearrange this into a standard quadratic equation:
\[ n^2 + n - 6 = 0 \]

### Step 4: Factor the quadratic equation
Now we need to factor the quadratic equation:
\[ n^2 + 3n - 2n - 6 = 0 \]
This can be factored as:
\[ (n + 3)(n - 2) = 0 \]

### Step 5: Solve for \(n\)
Setting each factor equal to zero gives us:
1. \(n + 3 = 0 \implies n = -3\)
2. \(n - 2 = 0 \implies n = 2\)

### Step 6: Determine valid solutions
Since \(n\) must be a non-negative integer (as factorials of negative numbers are not defined), we discard \(n = -3\). Thus, the only valid solution is:
\[ n = 2 \]

### Final Answer
The value of \(n\) is \(2\).

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