Let n be a positive integer and
`(1+x)^(n)+C_(0)+C_(1)x+C_(2)x^(2)+C_(3)x^(3)+ . . .+C_(r)x^(r)+ . . .+C_(n-1)x^(n-1)+C_(n)x^(n)`
Where `C_(r)` stands for `.^(n)C_(r)`, then
Q. The value of `underset(r=0)overset(n)(sum)underset(s=0)overset(n)(sum),C_(r),C_(S)` is

Options

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

Answer

To solve the given problem, we need to evaluate the double summation:

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

where \( C_r = \binom{n}{r} \).

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial expansion of \( (1+x)^n \) gives us:
\[ (1+x)^n = C_0 + C_1 x + C_2 x^2 + \ldots + C_n x^n \]
where \( C_r = \binom{n}{r} \).

2. **Setting Up the Summation**:
We need to evaluate:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r C_s \]

3. **Rearranging the Summation**:
Notice that the double summation can be rewritten using the property of summation:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r C_s = \left( \sum_{r=0}^{n} C_r \right) \left( \sum_{s=0}^{n} C_s \right) \]

4. **Calculating the Individual Sums**:
From the binomial theorem, we know:
\[ \sum_{r=0}^{n} C_r = (1+1)^n = 2^n \]
Therefore, we have:
\[ \sum_{r=0}^{n} C_r = 2^n \]
and similarly,
\[ \sum_{s=0}^{n} C_s = 2^n \]

5. **Combining the Results**:
Now substituting back into our rearranged summation:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} C_r C_s = (2^n)(2^n) = 2^{2n} \]

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

### Conclusion:
The final answer is:
\[ \boxed{2^{2n}} \]