`sum_(r=1)^n r (n-r +1)` is equal to :

Options

A. `(n(n^2 -1))/6`
B. `(n(n+1)(n+2))/6`
C. `(n(n^2-1))/2`
D. `(n(n+1)(n+2))/2`

Answer

To solve the problem \( \sum_{r=1}^{n} r(n - r + 1) \), we can follow these steps:

### Step 1: Expand the expression inside the summation
We start by expanding the term \( r(n - r + 1) \):
\[ r(n - r + 1) = rn - r^2 + r \]
Thus, we can rewrite the summation as:
\[ \sum_{r=1}^{n} r(n - r + 1) = \sum_{r=1}^{n} (rn - r^2 + r) \]

### Step 2: Split the summation
Now we can split the summation into three separate sums:
\[ \sum_{r=1}^{n} (rn - r^2 + r) = \sum_{r=1}^{n} rn - \sum_{r=1}^{n} r^2 + \sum_{r=1}^{n} r \]

### Step 3: Calculate each summation
1. **Calculate \( \sum_{r=1}^{n} rn \)**:
Since \( n \) is a constant with respect to \( r \), we can factor it out:
\[ \sum_{r=1}^{n} rn = n \sum_{r=1}^{n} r = n \cdot \frac{n(n + 1)}{2} \]

2. **Calculate \( \sum_{r=1}^{n} r^2 \)**:
The formula for the sum of squares of the first \( n \) natural numbers is:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n + 1)(2n + 1)}{6} \]

3. **Calculate \( \sum_{r=1}^{n} r \)**:
The formula for the sum of the first \( n \) natural numbers is:
\[ \sum_{r=1}^{n} r = \frac{n(n + 1)}{2} \]

### Step 4: Substitute back into the equation
Now substituting these results back into our expression:
\[ \sum_{r=1}^{n} r(n - r + 1) = n \cdot \frac{n(n + 1)}{2} - \frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2} \]

### Step 5: Combine the terms
Now we can combine the terms:
\[ = \frac{n(n + 1)}{2} + \frac{n(n + 1)}{2} - \frac{n(n + 1)(2n + 1)}{6} \]
This simplifies to:
\[ = n(n + 1) - \frac{n(n + 1)(2n + 1)}{6} \]

### Step 6: Find a common denominator
To combine these terms, we need a common denominator:
\[ = \frac{6n(n + 1)}{6} - \frac{n(n + 1)(2n + 1)}{6} \]
\[ = \frac{6n(n + 1) - n(n + 1)(2n + 1)}{6} \]

### Step 7: Simplify the numerator
Now we can simplify the numerator:
\[ = \frac{n(n + 1)(6 - (2n + 1))}{6} \]
\[ = \frac{n(n + 1)(5 - 2n)}{6} \]

### Final Result
Thus, the final answer is:
\[ \sum_{r=1}^{n} r(n - r + 1) = \frac{n(n + 1)(5 - 2n)}{6} \]