`(1+x)=C_(0)+C_(1)x+….+C_(n)x^(n)` then the value of `sumsum_(0lerltslen)C_(r)C_(s)` is equal to :

Options

A. `1/2[2^(2n)-""^(2n)C_(n)]`
B. `1/4[2^(2n)-""^(2n)C_(n)]`
C. `1/2[2^(2n)+""^(2n)C_(n)]`
D. `1/2[2^(n)-""^(2n)C_(n)]`

Answer

To solve the problem, we need to evaluate the double summation \( \sum_{r=0}^{n} \sum_{s=0}^{n} C_r C_s \), where \( C_r \) and \( C_s \) are the binomial coefficients from the expansion of \( (1+x)^n \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial expansion of \( (1+x)^n \) is given by:
\[ (1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]
where \( C_k = \binom{n}{k} \).

2. **Setting Up the Double Summation**:
We need to evaluate:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r C_s \]
This can be rewritten using the property of summation:
\[ \left( \sum_{r=0}^{n} C_r \right) \left( \sum_{s=0}^{n} C_s \right) \]
This is because the double summation can be factored into the product of two single summations.

3. **Calculating the Single Summation**:
From the binomial theorem, we know:
\[ \sum_{r=0}^{n} C_r = (1+1)^n = 2^n \]
Therefore, both summations yield \( 2^n \):
\[ \sum_{r=0}^{n} C_r = 2^n \quad \text{and} \quad \sum_{s=0}^{n} C_s = 2^n \]

4. **Calculating the Product**:
Now substituting back into our expression:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r C_s = (2^n)(2^n) = 2^{2n} \]

5. **Final Result**:
Thus, the value of the double summation \( \sum_{r=0}^{n} \sum_{s=0}^{n} C_r C_s \) is:
\[ 2^{2n} \]