The sum `sum_(r=0)^n (r+1) (C_r)^2` is equal to :

Options

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

Answer

To solve the problem of finding the sum \( S = \sum_{r=0}^{n} (r+1) \binom{n}{r}^2 \), we can follow these steps:

### Step 1: Rewrite the Sum
We can express the term \( (r+1) \) as \( r + 1 \):
\[ S = \sum_{r=0}^{n} (r+1) \binom{n}{r}^2 = \sum_{r=0}^{n} r \binom{n}{r}^2 + \sum_{r=0}^{n} \binom{n}{r}^2 \]

### Step 2: Simplify the First Sum
The first part \( \sum_{r=0}^{n} r \binom{n}{r}^2 \) can be simplified using the identity \( r \binom{n}{r} = n \binom{n-1}{r-1} \):
\[ \sum_{r=0}^{n} r \binom{n}{r}^2 = n \sum_{r=1}^{n} \binom{n-1}{r-1} \binom{n}{r} = n \sum_{r=0}^{n-1} \binom{n-1}{r} \binom{n}{r+1} \]

### Step 3: Use Vandermonde's Identity
Using Vandermonde's identity, we have:
\[ \sum_{r=0}^{k} \binom{m}{r} \binom{n}{k-r} = \binom{m+n}{k} \]
Applying this identity, we get:
\[ \sum_{r=0}^{n-1} \binom{n-1}{r} \binom{n}{r+1} = \binom{2n-1}{n} \]
Thus,
\[ \sum_{r=0}^{n} r \binom{n}{r}^2 = n \binom{2n-1}{n} \]

### Step 4: Simplify the Second Sum
The second part \( \sum_{r=0}^{n} \binom{n}{r}^2 \) can be simplified using the identity:
\[ \sum_{r=0}^{n} \binom{n}{r}^2 = \binom{2n}{n} \]

### Step 5: Combine the Results
Now we can combine the results from Step 3 and Step 4:
\[ S = n \binom{2n-1}{n} + \binom{2n}{n} \]

### Final Result
Thus, the sum \( S \) can be expressed as:
\[ S = n \binom{2n-1}{n} + \binom{2n}{n} \]