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)+C_(r)C_(s))` is :

Options

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

Answer

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

\[ \sum_{0 \leq r < s \leq n} (r+s)(C_r + C_s + C_r C_s) \]

where \( C_r \) and \( C_s \) are the binomial coefficients defined as \( C_r = \binom{n}{r} \) and \( C_s = \binom{n}{s} \).

### Step 1: Break down the expression

We can rewrite the expression inside the summation:

\[ (r+s)(C_r + C_s + C_r C_s) = (r+s)C_r + (r+s)C_s + (r+s)C_r C_s \]

Thus, we can separate the summation into three parts:

\[ \sum_{0 \leq r < s \leq n} (r+s)C_r + \sum_{0 \leq r < s \leq n} (r+s)C_s + \sum_{0 \leq r < s \leq n} (r+s)C_r C_s \]

### Step 2: Evaluate the first two sums

For the first two sums, notice that due to symmetry, we can combine them:

\[ \sum_{0 \leq r < s \leq n} (r+s)C_r = \sum_{0 \leq r < s \leq n} rC_r + \sum_{0 \leq r < s \leq n} sC_s \]

Since \( r \) and \( s \) are interchangeable in the summation, we can write:

\[ \sum_{0 \leq r < s \leq n} (r+s)C_r = \sum_{0 \leq r < s \leq n} rC_r = \sum_{0 \leq s < r \leq n} sC_s \]

Thus, we have:

\[ 2\sum_{0 \leq r < s \leq n} rC_r \]

### Step 3: Evaluate the third sum

For the third sum, we can rewrite it as:

\[ \sum_{0 \leq r < s \leq n} (r+s)C_r C_s \]

This can be expanded as:

\[ \sum_{0 \leq r < s \leq n} rC_r C_s + \sum_{0 \leq r < s \leq n} sC_r C_s \]

Again, due to symmetry, we can combine these:

\[ 2\sum_{0 \leq r < s \leq n} rC_r C_s \]

### Step 4: Combine all parts

Now we can combine all the parts:

\[ \sum_{0 \leq r < s \leq n} (r+s)(C_r + C_s + C_r C_s) = 2\sum_{0 \leq r < s \leq n} rC_r + 2\sum_{0 \leq r < s \leq n} rC_r C_s \]

### Step 5: Use binomial theorem properties

Using the properties of binomial coefficients, we know:

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

And for the product of coefficients:

\[ \sum_{r=0}^{n} C_r C_s = 2^n \]

### Final Calculation

Thus, we can conclude that:

\[ \sum_{0 \leq r < s \leq n} rC_r = \frac{n \cdot 2^{n-1}}{2} = n \cdot 2^{n-2} \]

And the final value of the original expression is:

\[ \text{Final Value} = 2 \cdot n \cdot 2^{n-2} + 2 \cdot n \cdot 2^{n-2} = 4n \cdot 2^{n-2} \]

### Conclusion

Thus, the value of the expression is:

\[ \boxed{4n \cdot 2^{n-2}} \]