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

Options

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

Answer

To solve the problem, we need to evaluate the expression:

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

where \( C_r \) is the binomial coefficient \( C(n, r) \).

### Step 1: Rewrite the Double Summation
We can separate the double summation into two parts:

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

### Step 2: Evaluate Each Part
Now, we can evaluate each part separately.

1. **First Part:**
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} r C_r C_s = \sum_{r=0}^{n} r C_r \sum_{s=0}^{n} C_s \]
The inner sum \( \sum_{s=0}^{n} C_s = 2^n \) (the sum of the binomial coefficients).

Thus, we have:
\[ \sum_{r=0}^{n} r C_r \cdot 2^n \]

2. **Second Part:**
Similarly,
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} s C_r C_s = \sum_{s=0}^{n} s C_s \sum_{r=0}^{n} C_r = 2^n \sum_{s=0}^{n} s C_s \]

### Step 3: Evaluate \( \sum_{r=0}^{n} r C_r \)
We know that:
\[ \sum_{r=0}^{n} r C_r = n \cdot 2^{n-1} \]
This is derived from the identity that states \( r C_r = n C_{r-1} \).

### Step 4: Combine the Results
Now, substituting back into our expression, we have:

1. From the first part:
\[ n \cdot 2^{n-1} \cdot 2^n = n \cdot 2^{2n-1} \]

2. From the second part:
\[ 2^n \cdot n \cdot 2^{n-1} = n \cdot 2^{2n-1} \]

### Step 5: Final Result
Adding both parts together:
\[ \sum_{r=0}^{n} \sum_{s=0}^{n} (r+s) C_r C_s = n \cdot 2^{2n-1} + n \cdot 2^{2n-1} = 2n \cdot 2^{2n-1} = n \cdot 2^{2n} \]

Thus, the final answer is:
\[ \boxed{n \cdot 2^{2n}} \]