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^(2n-3)+1/2""^(2n-2)C_(n-1)]`
B. `[2^(2n-3)-1/2""^(2n-2)C_(n-1)]`
C. `n^(2)[2^(2n-3)-1/2""^(2n-2)C_(n-1)]`
D. None of these

Answer

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

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

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

### Step 1: Understanding the Double Summation
The double summation can be interpreted as the sum of products of binomial coefficients multiplied by the indices \( r \) and \( s \).

### Step 2: Rewrite the Double Summation
We can rewrite the double summation as follows:

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

This allows us to separate the summation into two parts, one for \( r \) and one for \( s \).

### Step 3: Evaluate Each Summation
We know from combinatorial identities that:

\[ \sum_{r=0}^{n} r \cdot C_r = n \cdot 2^{n-1} \]

This is derived from the fact that \( r \cdot C_r = n \cdot C_{r-1} \), and thus the sum can be rewritten using the binomial theorem.

### Step 4: Substitute Back into the Expression
Now substituting back into our expression, we have:

\[ \sum_{r=0}^{n} r \cdot C_r \cdot \sum_{s=0}^{n} s \cdot C_s = (n \cdot 2^{n-1}) \cdot (n \cdot 2^{n-1}) = n^2 \cdot 2^{2(n-1)} \]

### Step 5: Final Result
Thus, the value of the double summation is:

\[ \sum_{r=0}^{n} \sum_{s=0}^{n} r \cdot s \cdot C_r \cdot C_s = n^2 \cdot 2^{2(n-1)} \]