The sum of the series ` sum_(r=0) ^(n) ""^(2n)C_(r), ` is

Options

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

Explanation

To find the sum of the series \( \sum_{r=0}^{n} \binom{2n}{r} \), we can use the Binomial Theorem. Here’s a step-by-step solution:

### Step 1: Understanding the Binomial Theorem
The Binomial Theorem states that:
\[ (x + y)^n = \sum_{r=0}^{n} \binom{n}{r} x^{n-r} y^r \]
For our case, we will use \( x = 1 \) and \( y = 1 \) to find the sum of the coefficients.

### Step 2: Applying the Binomial Theorem
We can express \( (1 + 1)^{2n} \) using the Binomial Theorem:
\[ (1 + 1)^{2n} = \sum_{r=0}^{2n} \binom{2n}{r} 1^{2n-r} 1^r = \sum_{r=0}^{2n} \binom{2n}{r} \]
This simplifies to:
\[ 2^{2n} = \sum_{r=0}^{2n} \binom{2n}{r} \]

### Step 3: Splitting the Sum
The sum \( \sum_{r=0}^{2n} \binom{2n}{r} \) can be split into two parts:
\[ \sum_{r=0}^{2n} \binom{2n}{r} = \sum_{r=0}^{n} \binom{2n}{r} + \sum_{r=n+1}^{2n} \binom{2n}{r} \]
By the symmetry property of binomial coefficients, we know:
\[ \binom{2n}{r} = \binom{2n}{2n - r} \]
Thus, the second sum can be rewritten as:
\[ \sum_{r=n+1}^{2n} \binom{2n}{r} = \sum_{r=0}^{n-1} \binom{2n}{r} \]

### Step 4: Relating the Sums
From the above, we can conclude:
\[ \sum_{r=0}^{n} \binom{2n}{r} = \sum_{r=n+1}^{2n} \binom{2n}{r} \]
This means:
\[ 2 \sum_{r=0}^{n} \binom{2n}{r} = 2^{2n} \]

### Step 5: Finding the Required Sum
Now, we can express the sum we need:
\[ \sum_{r=0}^{n} \binom{2n}{r} = \frac{1}{2} \times 2^{2n} = 2^{2n - 1} \]

### Final Result
Thus, the sum of the series \( \sum_{r=0}^{n} \binom{2n}{r} \) is:
\[ \boxed{2^{2n - 1}} \]