If `(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+….+C_(n)x^(n)`, then the value of `sumsum_(0lerltslen)(C_(r)+C_(s))^(2)` is :

Options

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

Answer

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

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

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

### Step-by-Step Solution:

1. **Expand the Square**:
\[ (C_r + C_s)^2 = C_r^2 + C_s^2 + 2C_rC_s \]

2. **Set Up the Double Sum**:
Substitute the expansion into the double sum:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} (C_r + C_s)^2 = \sum_{r=0}^{n} \sum_{s=0}^{n} (C_r^2 + C_s^2 + 2C_rC_s) \]

3. **Separate the Sums**:
This can be separated into three distinct sums:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r^2 + \sum_{r=0}^{n} \sum_{s=0}^{n} C_s^2 + 2 \sum_{r=0}^{n} \sum_{s=0}^{n} C_rC_s \]
Notice that \( \sum_{s=0}^{n} C_s^2 \) is the same as \( \sum_{r=0}^{n} C_r^2 \), so we can simplify:
\[ = 2\sum_{r=0}^{n} C_r^2 + 2\left(\sum_{r=0}^{n} C_r\right)^2 \]

4. **Use Known Results**:
- The sum of the binomial coefficients is:
\[ \sum_{r=0}^{n} C_r = 2^n \]
- The sum of the squares of the binomial coefficients is given by:
\[ \sum_{r=0}^{n} C_r^2 = C_{2n} = \binom{2n}{n} \]

5. **Substitute the Results**:
Substitute these results back into the equation:
\[ = 2 \cdot \binom{2n}{n} + 2(2^n)^2 \]
\[ = 2 \cdot \binom{2n}{n} + 2 \cdot 4^n \]

6. **Final Expression**:
Thus, the final value of the double sum is:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} (C_r + C_s)^2 = 2 \cdot \binom{2n}{n} + 2 \cdot 4^n \]

### Conclusion:
The value of the double sum is:
\[ 2 \cdot \binom{2n}{n} + 2 \cdot 4^n \]