`(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. `n^2/2[2^(2n-2)-""^(2n-2)C_(n-1)]`
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 find the value of the double sum \( \sum_{r=0}^{n} \sum_{s=r}^{n} C_r C_s \), where \( C_r \) and \( C_s \) are the binomial coefficients.

### Step-by-Step Solution:

1. **Understanding the Double Sum**:
The expression \( \sum_{r=0}^{n} \sum_{s=r}^{n} C_r C_s \) indicates that for each fixed \( r \), we sum over \( s \) starting from \( r \) to \( n \).

2. **Rearranging the Summation**:
We can switch the order of summation. The limits for \( s \) will go from \( 0 \) to \( n \), and for each \( s \), \( r \) will go from \( 0 \) to \( s \):
\[ \sum_{r=0}^{n} \sum_{s=r}^{n} C_r C_s = \sum_{s=0}^{n} \sum_{r=0}^{s} C_r C_s \]

3. **Factor Out \( C_s \)**:
Since \( C_s \) does not depend on \( r \), we can factor it out of the inner sum:
\[ \sum_{s=0}^{n} C_s \sum_{r=0}^{s} C_r \]

4. **Using the Binomial Theorem**:
The inner sum \( \sum_{r=0}^{s} C_r \) is equal to \( 2^s \) (by the binomial theorem, which states that the sum of the coefficients of \( (1+x)^s \) is \( 2^s \)):
\[ \sum_{r=0}^{s} C_r = 2^s \]
Therefore, we have:
\[ \sum_{s=0}^{n} C_s 2^s \]

5. **Final Calculation**:
The sum \( \sum_{s=0}^{n} C_s 2^s \) is equal to \( (1+2)^n = 3^n \) (again by the binomial theorem):
\[ \sum_{s=0}^{n} C_s 2^s = 3^n \]

6. **Conclusion**:
Thus, the value of the double sum \( \sum_{r=0}^{n} \sum_{s=r}^{n} C_r C_s \) is:
\[ \sum_{r=0}^{n} \sum_{s=r}^{n} C_r C_s = 3^n \]

### Final Answer:
The value of \( \sum_{r=0}^{n} \sum_{s=r}^{n} C_r C_s \) is \( 3^n \).