`Sigma_(r=0)^(n)((r^(2))/(r+1)).^(n)C_(r)` is equal to

Options

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

Answer

To solve the problem \( \sum_{r=0}^{n} \frac{r^2}{r+1} \binom{n}{r} \), we will follow these steps:

### Step 1: Rewrite the Summation
We start with the original summation:
\[ \sum_{r=0}^{n} \frac{r^2}{r+1} \binom{n}{r} \]
We can rewrite \( r^2 \) as \( r^2 = (r-1)(r+1) + (r-1) \). Thus, we can express the summation as:
\[ \sum_{r=0}^{n} \frac{(r-1)(r+1) + (r-1)}{r+1} \binom{n}{r} \]
This simplifies to:
\[ \sum_{r=0}^{n} \left( (r-1) + 1 \right) \binom{n}{r} \]

### Step 2: Split the Summation
Now we can split the summation into two parts:
\[ \sum_{r=0}^{n} (r-1) \binom{n}{r} + \sum_{r=0}^{n} \binom{n}{r} \]
The second summation \( \sum_{r=0}^{n} \binom{n}{r} = 2^n \).

### Step 3: Simplify the First Summation
For the first summation, we can use the identity \( r \binom{n}{r} = n \binom{n-1}{r-1} \):
\[ \sum_{r=0}^{n} (r-1) \binom{n}{r} = \sum_{r=1}^{n} (r-1) \binom{n}{r} = \sum_{r=1}^{n} (r \binom{n}{r} - \binom{n}{r}) = n \sum_{r=1}^{n} \binom{n-1}{r-1} - \sum_{r=0}^{n} \binom{n}{r} \]
The first part simplifies to \( n \cdot 2^{n-1} \) and the second part is \( 2^n \):
\[ = n \cdot 2^{n-1} - 2^n \]

### Step 4: Combine the Results
Now we combine the results:
\[ \sum_{r=0}^{n} \frac{r^2}{r+1} \binom{n}{r} = n \cdot 2^{n-1} - 2^n + 2^n = n \cdot 2^{n-1} \]

### Final Result
Thus, the final result is:
\[ \sum_{r=0}^{n} \frac{r^2}{r+1} \binom{n}{r} = n \cdot 2^{n-1} \]