If `C_(0), C_(1) C_(2)` ….., denote the binomial coefficients
in the expansion of `(1 + x)^(n)` , then
`(C_(0))/(2) - (C_(1))/(3) + (C_(2))/(4)- (C_(3))/(5)+...+ (-1)^(n)(C_(n))/(n+2) = `

Options

A. `(1)/(n+1)`
B. `(1)/(n+2)`
C. `(1)/(n(n+1))`
D. `(1)/((n+1)(n+2))`

Explanation

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

\[ S = \frac{C_0}{2} - \frac{C_1}{3} + \frac{C_2}{4} - \frac{C_3}{5} + \ldots + (-1)^n \frac{C_n}{n+2} \]

where \( C_k = \binom{n}{k} \) are the binomial coefficients from the expansion of \( (1+x)^n \).

### Step 1: Write the binomial expansion
The binomial expansion of \( (1+x)^n \) is given by:

\[ (1+x)^n = \sum_{k=0}^{n} C_k x^k = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]

### Step 2: Substitute \( x = -x \)
To incorporate the alternating signs in the sum, we substitute \( x \) with \( -x \):

\[ (1-x)^n = \sum_{k=0}^{n} C_k (-x)^k = C_0 - C_1 x + C_2 x^2 - C_3 x^3 + \ldots + (-1)^n C_n x^n \]

### Step 3: Multiply by \( x \)
Now, we multiply the entire equation by \( x \):

\[ x(1-x)^n = xC_0 - xC_1 x + xC_2 x^2 - xC_3 x^3 + \ldots + (-1)^n C_n x^{n+1} \]

This simplifies to:

\[ x(1-x)^n = C_0 x - C_1 x^2 + C_2 x^3 - C_3 x^4 + \ldots + (-1)^n C_n x^{n+1} \]

### Step 4: Integrate from 0 to 1
Next, we integrate the expression from 0 to 1:

\[ \int_0^1 x(1-x)^n \, dx \]

Using the integration by parts or the known result for the beta function, we find:

\[ \int_0^1 x(1-x)^n \, dx = \frac{1}{(n+1)(n+2)} \]

### Step 5: Relate the integral to the sum
The integral we computed corresponds to the sum we are trying to evaluate:

\[ \int_0^1 \left( C_0 x - C_1 \frac{x^2}{2} + C_2 \frac{x^3}{3} - C_3 \frac{x^4}{4} + \ldots + (-1)^n C_n \frac{x^{n+1}}{n+2} \right) dx \]

This gives us:

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

### Final Answer
Thus, the value of the sum is:

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