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

Options

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

Answer

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

### Step 1: Expand the summation
We start with the expression:
\[ \sum_{r=1}^{n} r(n-r) \]
This can be rewritten as:
\[ \sum_{r=1}^{n} (nr - r^2) \]
Now we can separate the summation:
\[ = n \sum_{r=1}^{n} r - \sum_{r=1}^{n} r^2 \]

### Step 2: Use the formula for the sum of the first \( n \) natural numbers
The formula for the sum of the first \( n \) natural numbers is:
\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]
Substituting this into our expression gives:
\[ = n \left(\frac{n(n+1)}{2}\right) - \sum_{r=1}^{n} r^2 \]

### Step 3: Use the formula for the sum of the squares of the first \( n \) natural numbers
The formula for the sum of the squares of the first \( n \) natural numbers is:
\[ \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \]
Now substituting this into our expression, we have:
\[ = n \left(\frac{n(n+1)}{2}\right) - \frac{n(n+1)(2n+1)}{6} \]

### Step 4: Combine the terms
To combine these terms, we need a common denominator. The common denominator between 2 and 6 is 6:
\[ = \frac{3n(n(n+1))}{6} - \frac{n(n+1)(2n+1)}{6} \]
Now we can combine the fractions:
\[ = \frac{3n(n(n+1)) - n(n+1)(2n+1)}{6} \]

### Step 5: Factor out common terms
Notice that \( n(n+1) \) is common in both terms:
\[ = \frac{n(n+1) \left(3n - (2n + 1)\right)}{6} \]
Simplifying the expression inside the parentheses:
\[ = \frac{n(n+1)(3n - 2n - 1)}{6} \]
\[ = \frac{n(n+1)(n - 1)}{6} \]

### Final Result
Thus, the result of the summation \( \sum_{r=1}^{n} r(n-r) \) is:
\[ \frac{n(n+1)(n-1)}{6} \]