`sum_(r=1)^(n) r^(2)-sum_(r=1)^(n) sum_(r=1)^(n) ` is equal to

Options

A. 0
B. `(1)/(2)(underset(r=1)overset(n)sumr^(2)+underset(r=1)overset(n)sumr)`
C. `(1)/(2){underset(r=1)overset(n)sumr^(2)-underset(r=1)overset(n)sumr}`
D. none of these

Explanation

To solve the expression \( \sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} \sum_{r=1}^{n} r \), we will break down the problem step by step.

### Step 1: Understanding the Summations
The first part of the expression is \( \sum_{r=1}^{n} r^2 \), which is the sum of squares of the first \( n \) natural numbers. The second part is \( \sum_{r=1}^{n} \sum_{r=1}^{n} r \), which represents the sum of \( r \) taken \( n \) times.

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

### Step 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: Calculate \( \sum_{r=1}^{n} \sum_{r=1}^{n} r \)
Since \( \sum_{r=1}^{n} r \) is a single sum, when we sum it \( n \) times, we get:
\[ \sum_{r=1}^{n} \sum_{r=1}^{n} r = n \cdot \sum_{r=1}^{n} r = n \cdot \frac{n(n+1)}{2} = \frac{n^2(n+1)}{2} \]

### Step 5: Substitute Back into the Original Expression
Now we substitute the results from Steps 2 and 4 back into the original expression:
\[ \sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} \sum_{r=1}^{n} r = \frac{n(n+1)(2n+1)}{6} - \frac{n^2(n+1)}{2} \]

### Step 6: Simplify the Expression
To simplify, we need a common denominator. The common denominator between 6 and 2 is 6. Thus, we rewrite the second term:
\[ \frac{n^2(n+1)}{2} = \frac{3n^2(n+1)}{6} \]
Now we can combine the two fractions:
\[ \frac{n(n+1)(2n+1)}{6} - \frac{3n^2(n+1)}{6} = \frac{n(n+1)(2n+1 - 3n)}{6} = \frac{n(n+1)(-n+1)}{6} \]

### Final Answer
Thus, the final expression simplifies to:
\[ \frac{n(n+1)(1-n)}{6} \]