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

Options

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

Explanation

To find the sum of the series \( S = \sum_{r=0}^{n} r \cdot \binom{2n}{r} \), we can follow these steps:

### Step 1: Rewrite the Series
We start with the series:
\[ S = \sum_{r=0}^{n} r \cdot \binom{2n}{r} \]
We can express \( r \cdot \binom{2n}{r} \) in terms of binomial coefficients:
\[ r \cdot \binom{2n}{r} = 2n \cdot \binom{2n-1}{r-1} \]
Thus, we can rewrite the series \( S \) as:
\[ S = \sum_{r=1}^{n} 2n \cdot \binom{2n-1}{r-1} \]

### Step 2: Factor Out the Constant
Since \( 2n \) is a constant with respect to the summation, we can factor it out:
\[ S = 2n \sum_{r=1}^{n} \binom{2n-1}{r-1} \]

### Step 3: Change the Index of Summation
Next, we change the index of summation by letting \( k = r - 1 \). When \( r = 1 \), \( k = 0 \), and when \( r = n \), \( k = n - 1 \):
\[ S = 2n \sum_{k=0}^{n-1} \binom{2n-1}{k} \]

### Step 4: Use the Binomial Theorem
According to the binomial theorem, the sum of the binomial coefficients gives:
\[ \sum_{k=0}^{m} \binom{m}{k} = 2^m \]
Thus, we can apply this to our sum:
\[ \sum_{k=0}^{2n-1} \binom{2n-1}{k} = 2^{2n-1} \]
However, we only need the sum from \( k = 0 \) to \( n-1 \). This is half of the total sum of the binomial coefficients:
\[ \sum_{k=0}^{n-1} \binom{2n-1}{k} = \frac{1}{2} \cdot 2^{2n-1} = 2^{2n-2} \]

### Step 5: Substitute Back into the Expression for \( S \)
Now we can substitute this result back into our expression for \( S \):
\[ S = 2n \cdot 2^{2n-2} \]

### Step 6: Simplify the Expression
Finally, we simplify the expression:
\[ S = n \cdot 2^{2n-1} \]

### Final Answer
Therefore, the sum of the series is:
\[ \boxed{n \cdot 2^{2n-1}} \]