Solve for n :
` (n+1) ! = 56 (n-1) ! `

Answer

To solve the equation \( (n+1)! = 56(n-1)! \), we can follow these steps:

### Step 1: Rewrite the factorials
We know that:
\[ (n+1)! = (n+1) \cdot n \cdot (n-1)! \]
Substituting this into the equation gives:
\[ (n+1) \cdot n \cdot (n-1)! = 56(n-1)! \]

### Step 2: Cancel out the common factorial
Since \( (n-1)! \) appears on both sides of the equation, we can cancel it out (assuming \( n \neq 1 \)):
\[ (n+1) \cdot n = 56 \]

### Step 3: Expand the equation
Expanding the left side, we have:
\[ n^2 + n = 56 \]

### Step 4: Rearrange the equation
Rearranging gives us a standard quadratic equation:
\[ n^2 + n - 56 = 0 \]

### Step 5: Factor the quadratic equation
To factor the quadratic, we look for two numbers that multiply to \(-56\) and add to \(1\). The numbers \(8\) and \(-7\) fit:
\[ (n + 8)(n - 7) = 0 \]

### Step 6: Solve for \(n\)
Setting each factor to zero gives us:
\[ n + 8 = 0 \quad \Rightarrow \quad n = -8 \quad (\text{not valid since } n \text{ must be a non-negative integer}) \]
\[ n - 7 = 0 \quad \Rightarrow \quad n = 7 \]

### Final Answer
Thus, the solution is:
\[ n = 7 \]