If n `in` N, then `sum_(r=0)^(n) (-1)^(n) (""^(n)C_(r))/(""^(r+2)C_(r))` is equal to .

Options

A. `(3!)/(2(n-3))`
B. `(3!)/(2(n+3))`
C. `(3!)/((n+3))`
D. None of these

Answer

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{r=0}^{n} (-1)^n \frac{nC_r}{(r+2)C_r} \]

### Step 1: Rewrite the binomial coefficients
We start by rewriting the binomial coefficients in the summation. The binomial coefficient \( nC_r \) can be expressed as:

\[ nC_r = \frac{n!}{r!(n-r)!} \]

Similarly, the binomial coefficient \( (r+2)C_r \) can be expressed as:

\[ (r+2)C_r = \frac{(r+2)!}{r! \cdot 2!} = \frac{(r+2)(r+1)}{2} \]

### Step 2: Substitute the coefficients into the sum
Substituting these expressions into the sum gives:

\[ S = \sum_{r=0}^{n} (-1)^n \frac{\frac{n!}{r!(n-r)!}}{\frac{(r+2)(r+1)}{2}} = \sum_{r=0}^{n} (-1)^n \frac{2n!}{(r+2)(r+1)r!(n-r)!} \]

### Step 3: Factor out constants
Since \( (-1)^n \) and \( 2n! \) are constants with respect to the summation, we can factor them out:

\[ S = 2(-1)^n n! \sum_{r=0}^{n} \frac{1}{(r+2)(r+1)} \frac{1}{(n-r)!} \]

### Step 4: Simplify the summation
Now, we simplify the summation:

\[ \sum_{r=0}^{n} \frac{1}{(r+2)(r+1)} \frac{1}{(n-r)!} \]

This can be rewritten as:

\[ \sum_{r=0}^{n} \frac{1}{(r+2)(r+1)} \cdot \frac{1}{(n-r)!} \]

### Step 5: Change the summation index
To simplify the summation further, we can change the index of summation by letting \( k = n - r \). Then \( r = n - k \), and as \( r \) goes from \( 0 \) to \( n \), \( k \) goes from \( n \) to \( 0 \):

\[ S = 2(-1)^n n! \sum_{k=0}^{n} \frac{1}{(n-k+2)(n-k+1)} \cdot \frac{1}{k!} \]

### Step 6: Recognize the pattern
Now we can recognize that this summation can be related to the binomial theorem. The sum can be evaluated using the binomial expansion, and we can find that:

\[ \sum_{k=0}^{n} \frac{1}{(n-k+2)(n-k+1)} \cdot \frac{1}{k!} = \frac{1}{(n+2)!} \]

### Step 7: Final expression
Putting it all together, we have:

\[ S = 2(-1)^n n! \cdot \frac{1}{(n+2)!} \]

This simplifies to:

\[ S = \frac{2(-1)^n}{(n+2)(n+1)} \]

### Final Answer
Thus, the final answer is:

\[ \frac{2(-1)^n}{(n+2)(n+1)} \]