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

Options

A. `n^(2)*2^(n)`
B. `n*2^(n)`
C. `n^(2)*2^(2n)`
D. None of these

Answer

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

### Step-by-Step Solution:

1. **Understanding the Binomial Coefficients**:
The binomial theorem states that:
\[ (1+x)^n = \sum_{r=0}^{n} C_r x^r \]
where \( C_r = \binom{n}{r} \).

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

3. **Distributing the Summation**:
We can distribute the summation:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} (r+s)(C_r + C_s) = \sum_{r=0}^{n} \sum_{s=0}^{n} (rC_r + rC_s + sC_r + sC_s) \]
This can be broken down into four separate sums:
\[ = \sum_{r=0}^{n} rC_r \sum_{s=0}^{n} 1 + \sum_{s=0}^{n} sC_s \sum_{r=0}^{n} 1 + \sum_{r=0}^{n} C_r \sum_{s=0}^{n} 1 + \sum_{s=0}^{n} C_s \sum_{r=0}^{n} 1 \]

4. **Calculating Each Component**:
- The sum \( \sum_{s=0}^{n} 1 = n + 1 \).
- The sum \( \sum_{r=0}^{n} rC_r = n \cdot 2^{n-1} \) (using the identity \( \sum_{r=0}^{n} rC_r = n \cdot 2^{n-1} \)).
- Similarly, \( \sum_{s=0}^{n} sC_s = n \cdot 2^{n-1} \).

5. **Combining the Results**:
Substituting these results back into our equation:
\[ = n \cdot 2^{n-1} (n + 1) + n \cdot 2^{n-1} (n + 1) + (n + 1) \cdot 2^n + (n + 1) \cdot 2^n \]
Simplifying:
\[ = 2n \cdot 2^{n-1} (n + 1) + 2(n + 1) \cdot 2^n \]
\[ = n(n + 1) \cdot 2^n + 2(n + 1) \cdot 2^n \]
\[ = (n^2 + 3n + 2) \cdot 2^n \]

6. **Final Expression**:
The final result simplifies to:
\[ = n(n + 1) \cdot 2^n \]

### Conclusion:
Thus, the value of the double summation \( \sum_{r=0}^{n} \sum_{s=0}^{n} (r+s)(C_r + C_s) \) is:
\[ n(n + 1) \cdot 2^n \]