if `(n+2)! =2550xxn!`, find the value of n.

Explanation

To solve the equation \((n+2)! = 2550 \times n!\), we will follow these steps:

### Step 1: Rewrite the factorial equation
We start with the equation:
\[ (n + 2)! = 2550 \times n! \]
We can express \((n + 2)!\) in terms of \(n!\):
\[ (n + 2)! = (n + 2)(n + 1)(n!) \]
Substituting this into the equation gives us:
\[ (n + 2)(n + 1)(n!) = 2550 \times n! \]

### Step 2: Cancel \(n!\)
Since \(n!\) appears on both sides of the equation, we can cancel it (assuming \(n! \neq 0\)):
\[ (n + 2)(n + 1) = 2550 \]

### Step 3: Expand the left side
Now we expand the left side:
\[ n^2 + 3n + 2 = 2550 \]

### Step 4: Rearrange into standard quadratic form
Next, we rearrange the equation to set it to zero:
\[ n^2 + 3n + 2 - 2550 = 0 \]
This simplifies to:
\[ n^2 + 3n - 2548 = 0 \]

### Step 5: Factor the quadratic equation
Now we will factor the quadratic equation. We need two numbers that multiply to \(-2548\) and add to \(3\). The factors are \(52\) and \(-49\):
\[ (n + 52)(n - 49) = 0 \]

### Step 6: Solve for \(n\)
Setting each factor to zero gives us:
1. \(n + 52 = 0 \Rightarrow n = -52\)
2. \(n - 49 = 0 \Rightarrow n = 49\)

### Step 7: Determine the valid solution
Since \(n\) must be a non-negative integer (as it represents a factorial), we discard \(n = -52\) and accept:
\[ n = 49 \]

### Final Answer
The value of \(n\) is:
\[ \boxed{49} \]

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