If `C_0,C_1,C_2,.......,C_n` denote the binomial coefficients in the expansion of `(1+1)^n,` then `sum_(0ler) sum_(< s le n) (C_r+C_s)`

Options

A. `2^(n)`
B. `2^(n-1)`
C. `n.2^(n)`
D. none of these

Explanation

To solve the problem, we need to evaluate the double summation of the binomial coefficients \( C_r + C_s \) where \( r \) and \( s \) range from \( 0 \) to \( n \).

### Step-by-Step Solution:

1. **Understanding the Double Summation**:
The expression we need to evaluate is:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} (C_r + C_s) \]
This can be split into two separate summations:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r + \sum_{r=0}^{n} \sum_{s=0}^{n} C_s \]

2. **Evaluating Each Summation**:
The first summation can be simplified:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r = \sum_{r=0}^{n} C_r \cdot (n + 1) \]
because for each \( C_r \), there are \( n + 1 \) terms of \( C_s \) (from \( s = 0 \) to \( s = n \)).

Thus, we have:
\[ \sum_{r=0}^{n} C_r = 2^n \quad \text{(from the binomial theorem)} \]
Therefore:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r = (n + 1) \cdot 2^n \]

3. **Evaluating the Second Summation**:
Similarly, for the second summation:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_s = \sum_{s=0}^{n} C_s \cdot (n + 1) = (n + 1) \cdot 2^n \]

4. **Combining the Results**:
Now, we combine both results:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} (C_r + C_s) = (n + 1) \cdot 2^n + (n + 1) \cdot 2^n = 2(n + 1) \cdot 2^n \]

5. **Final Answer**:
Thus, the final result simplifies to:
\[ = (n + 1) \cdot 2^{n + 1} \]

### Conclusion:
The value of the double summation \( \sum_{r=0}^{n} \sum_{s=0}^{n} (C_r + C_s) \) is:
\[ (n + 1) \cdot 2^{n + 1} \]