`Sigma_(r=0)^(n)(n-r)(.^(n)C_(r))^(2)` is equal to

Options

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

Answer

To solve the problem \( \sum_{r=0}^{n} (n-r) \binom{n}{r}^2 \), we can follow these steps:

### Step 1: Rewrite the summation
We start with the expression:
\[ \sum_{r=0}^{n} (n-r) \binom{n}{r}^2 \]
This can be rewritten as:
\[ n \sum_{r=0}^{n} \binom{n}{r}^2 - \sum_{r=0}^{n} r \binom{n}{r}^2 \]

### Step 2: Evaluate the first summation
The first summation \( \sum_{r=0}^{n} \binom{n}{r}^2 \) can be evaluated using the identity:
\[ \sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n} \]
Thus, we have:
\[ n \sum_{r=0}^{n} \binom{n}{r}^2 = n \binom{2n}{n} \]

### Step 3: Evaluate the second summation
For the second summation \( \sum_{r=0}^{n} r \binom{n}{r}^2 \), we can use the identity:
\[ r \binom{n}{r} = n \binom{n-1}{r-1} \]
This allows us to rewrite the summation as:
\[ \sum_{r=0}^{n} r \binom{n}{r}^2 = n \sum_{r=1}^{n} \binom{n-1}{r-1} \binom{n}{r} = n \sum_{s=0}^{n-1} \binom{n-1}{s} \binom{n}{s+1} \]
Using the Vandermonde identity, we can evaluate this as:
\[ n \binom{2n-1}{n} \]

### Step 4: Combine results
Now we can combine our results:
\[ \sum_{r=0}^{n} (n-r) \binom{n}{r}^2 = n \binom{2n}{n} - n \binom{2n-1}{n} \]
Factoring out \( n \):
\[ = n \left( \binom{2n}{n} - \binom{2n-1}{n} \right) \]

### Step 5: Simplify the expression
Using the identity \( \binom{2n}{n} = \binom{2n-1}{n} + \binom{2n-1}{n-1} \), we can simplify:
\[ \binom{2n}{n} - \binom{2n-1}{n} = \binom{2n-1}{n-1} \]
Thus, we have:
\[ \sum_{r=0}^{n} (n-r) \binom{n}{r}^2 = n \binom{2n-1}{n-1} \]

### Final Answer
The final result is:
\[ \sum_{r=0}^{n} (n-r) \binom{n}{r}^2 = n \binom{2n-1}{n-1} \]