If `C_(0), C_(1), C_(2), …, C_(n)` denote the binomial
coefficients in the expansion of `(1 + x)^(n) ` , then `sum_(0 ler )^(n)sum_(lt s len)^(n)C_(r)C_(s)`=.

Options

A. `2^(2n)-""^(2n)C_(n)`
B. `""^(2n)C_(n)-2^(2n)`
C. `(1)/(2)(2""^(2n)-""^(2n)C_(n))`
D. `(1)/(2)(2""^(2n)+""^(2n)C_(n))`

Explanation

To solve the problem, we need to find the value of the expression:

\[ \sum_{r=0}^{n} \sum_{s=r+1}^{n} C_r C_s \]

where \( C_k \) represents the binomial coefficients from the expansion of \( (1 + x)^n \).

### Step 1: Understanding the Double Summation
The double summation can be interpreted as summing the products of binomial coefficients \( C_r \) and \( C_s \) for all pairs \( (r, s) \) such that \( 0 \leq r < s \leq n \).

### Step 2: Rewriting the Double Summation
We can express the double summation as follows:

\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r C_s - \sum_{r=0}^{n} C_r C_r \]

The first term sums over all pairs \( (r, s) \), while the second term subtracts the cases where \( r = s \) (i.e., the diagonal elements).

### Step 3: Evaluating the First Term
The first term can be simplified using the binomial theorem:

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

Thus,

\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r C_s = \left( \sum_{r=0}^{n} C_r \right)^2 = (2^n)^2 = 2^{2n} \]

### Step 4: Evaluating the Second Term
The second term involves the sum of squares of the binomial coefficients:

\[ \sum_{r=0}^{n} C_r^2 = C_{2n}^n \]

This is a known result from combinatorics, which states that the sum of the squares of the binomial coefficients is equal to the central binomial coefficient of \( 2n \).

### Step 5: Putting It All Together
Now we can substitute back into our expression:

\[ \sum_{r=0}^{n} \sum_{s=r+1}^{n} C_r C_s = \frac{1}{2} \left( \sum_{r=0}^{n} \sum_{s=0}^{n} C_r C_s - \sum_{r=0}^{n} C_r^2 \right) \]

Substituting the values we found:

\[ \sum_{r=0}^{n} \sum_{s=r+1}^{n} C_r C_s = \frac{1}{2} \left( 2^{2n} - C_{2n}^n \right) \]

### Final Answer
Thus, the required summation is:

\[ \sum_{r=0}^{n} \sum_{s=r+1}^{n} C_r C_s = \frac{1}{2} \left( 2^{2n} - C_{2n}^n \right) \]