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

Options

A. `(n+1)""^(2n)C_(n)`
B. `(n+1)""^(2n)C_(n)+2^(n)`
C. `(n+1)""^(2n)C_(n)`
D. `(n-1)""^(2n)C_(n)+2^(2n)`

Answer

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

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

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

### Step 1: Expand the Expression
We start by expanding the square:

\[ (C_r + C_s)^2 = C_r^2 + C_s^2 + 2C_r C_s \]

### Step 2: Set Up the Double Summation
Now, we can express the double summation:

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

### Step 3: Simplify the Summation
Notice that the summation of \( C_r^2 \) and \( C_s^2 \) can be combined because they are symmetric:

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

Thus, we can write:

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

### Step 4: Calculate \( \sum_{0 \leq r \leq n} C_r^2 \)
Using the identity for the sum of squares of binomial coefficients:

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

This means:

\[ \sum_{0 \leq r \leq s \leq n} C_r^2 = \frac{1}{2} \sum_{r=0}^{n} C_r^2 = \frac{1}{2} C_{2n} \]

### Step 5: Evaluate \( \sum_{0 \leq r \leq s \leq n} C_r C_s \)
For the term \( 2 \sum_{0 \leq r \leq s \leq n} C_r C_s \), we can express it as:

\[ \sum_{0 \leq r, s \leq n} C_r C_s = \left( \sum_{r=0}^{n} C_r \right)^2 = (2^n)^2 = 4^n \]

However, since we only want the case where \( r \leq s \), we have:

\[ \sum_{0 \leq r \leq s \leq n} C_r C_s = \frac{1}{2} \sum_{0 \leq r, s \leq n} C_r C_s = \frac{1}{2} \cdot 4^n \]

### Step 6: Combine All Parts
Now we can combine all parts:

\[ \sum_{0 \leq r \leq s \leq n} (C_r + C_s)^2 = 2 \cdot \frac{1}{2} C_{2n} + 2 \cdot \frac{1}{2} \cdot 4^n \]

This simplifies to:

\[ C_{2n} + 4^n \]

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

\[ \sum_{0 \leq r \leq s \leq n} (C_r + C_s)^2 = C_{2n} + 4^n \]