Let n be a positive integer and
`(1+x)^(n)=C_(0)+C_(1)x+C_(2)x^(2)+C_(3)x^(3)+ . . .+C_(r)x^(r)+ . . .+C_(n-1)x^(n-1)+C_(n)x^(n)`
Where `C_(r)` stands for `.^(n)C_(r)`, then
Q. The values of `underset(r=0)overset(n)(sum)underset(s=0)overset(n)(sum)(C_(r)+C_(S))` is

Options

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

Answer

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

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

where \( C_r = \binom{n}{r} \).

### Step 1: Rewrite the Double Summation
We can rewrite the double summation as follows:

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

### Step 2: Simplify the Inner Summation
Since \( C_r \) does not depend on \( s \), we can factor it out of the inner summation:

\[ \sum_{s=0}^{n} C_s = \sum_{s=0}^{n} \binom{n}{s} \]

The sum of the binomial coefficients from \( 0 \) to \( n \) is known to equal \( 2^n \):

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

Thus, we can rewrite our expression:

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

### Step 3: Combine the Results
Now, substituting back into our original expression, we have:

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

### Step 4: Final Result
Thus, the final answer is:

\[ 2^{n+1} (n+1) \]