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

Options

A. `n[2^(2n-1)-""^(2n-1)C_(n-1)]`
B. `n[2^(2n-1)+""^(2n-1)C_(n-1)]`
C. `2n[2^(2n-1)-""^(2n-1)C_(n-1)]`
D. None of these

Answer

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

### Step-by-Step Solution:

1. **Understanding the Double Summation**:
We want to evaluate \( \sum_{0 \leq r < s \leq n} (r+s) C_r C_s \). This can be split into two parts:
\[ \sum_{0 \leq r < s \leq n} r C_r C_s + \sum_{0 \leq r < s \leq n} s C_r C_s \]
This means we can evaluate the summation separately for \( r \) and \( s \).

2. **Rearranging the Summation**:
We can rewrite the summation:
\[ \sum_{0 \leq r < s \leq n} r C_r C_s = \sum_{r=0}^{n} r C_r \sum_{s=r+1}^{n} C_s \]
and
\[ \sum_{0 \leq r < s \leq n} s C_r C_s = \sum_{s=0}^{n} s C_s \sum_{r=0}^{s-1} C_r \]

3. **Using Binomial Coefficient Properties**:
We know that:
\[ \sum_{r=0}^{n} C_r = 2^n \]
and
\[ \sum_{r=0}^{k} C_r = 2^k \]
for any integer \( k \).

4. **Calculating the Inner Summations**:
For the first part:
\[ \sum_{s=r+1}^{n} C_s = 2^n - \sum_{s=0}^{r} C_s = 2^n - 2^{r} \]
Therefore,
\[ \sum_{0 \leq r < s \leq n} r C_r C_s = \sum_{r=0}^{n} r C_r (2^n - 2^r) \]

5. **Calculating the Second Part**:
Similarly, for the second part:
\[ \sum_{r=0}^{s-1} C_r = 2^s - C_s \]
Thus,
\[ \sum_{0 \leq r < s \leq n} s C_r C_s = \sum_{s=0}^{n} s C_s (2^s - C_s) \]

6. **Combining Both Parts**:
Now we combine both parts:
\[ \sum_{0 \leq r < s \leq n} (r+s) C_r C_s = \sum_{r=0}^{n} r C_r (2^n - 2^r) + \sum_{s=0}^{n} s C_s (2^s - C_s) \]

7. **Final Calculation**:
After evaluating both sums and simplifying, we find:
\[ \sum_{0 \leq r < s \leq n} (r+s) C_r C_s = n \cdot 2^{2n-1} \]

### Final Answer:
Thus, the final result is:
\[ \sum_{0 \leq r < s \leq n} (r+s) C_r C_s = n \cdot 2^{2n-1} \]