what is ` ((n+2)! +(n+1)(n-1)!)/((n+1) (n-1)!)` equal to?

Options

A. 1
B. Always an odd integer
C. A perfect square
D. None of the above

Answer

To solve the expression \(\frac{(n+2)! + (n+1)(n-1)!}{(n+1)(n-1)!}\), we will follow these steps:

### Step 1: Expand the factorials
We start by expanding \((n+2)!\):
\[ (n+2)! = (n+2)(n+1)(n)(n-1)! \]
Now we can rewrite the expression:
\[ \frac{(n+2)(n+1)(n)(n-1)! + (n+1)(n-1)!}{(n+1)(n-1)!} \]

### Step 2: Factor out the common term
In the numerator, we can factor out \((n+1)(n-1)!\):
\[ = \frac{(n+1)(n-1)! \left[(n+2)n + 1\right]}{(n+1)(n-1)!} \]

### Step 3: Cancel the common terms
Now, we can cancel \((n+1)(n-1)!\) from the numerator and the denominator:
\[ = (n+2)n + 1 \]

### Step 4: Simplify the expression
Now we simplify the expression:
\[ = n^2 + 2n + 1 \]

### Step 5: Recognize the perfect square
The expression \(n^2 + 2n + 1\) can be recognized as:
\[ = (n+1)^2 \]

### Final Answer
Thus, the final result is:
\[ (n+1)^2 \]