`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 problem, we need to evaluate the expression:

\[ \sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} \sum_{r=1}^{n} r \]

### Step 1: Evaluate the first summation \(\sum_{r=1}^{n} r^2\)

We know the formula for the sum of squares of the first \(n\) natural numbers:

\[ \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6} \]

### Step 2: Evaluate the second summation \(\sum_{r=1}^{n} r\)

The formula for the sum of the first \(n\) natural numbers is:

\[ \sum_{r=1}^{n} r = \frac{n(n+1)}{2} \]

### Step 3: Substitute the values into the expression

Now, we substitute the values from Steps 1 and 2 into the original expression:

\[ \sum_{r=1}^{n} r^2 - \sum_{r=1}^{n} r = \frac{n(n+1)(2n+1)}{6} - \frac{n(n+1)}{2} \]

### Step 4: Simplify the expression

To simplify the expression, we need a common denominator. The common denominator between 6 and 2 is 6. Thus, we rewrite the second term:

\[ \frac{n(n+1)}{2} = \frac{3n(n+1)}{6} \]

Now, we can rewrite the expression:

\[ \frac{n(n+1)(2n+1)}{6} - \frac{3n(n+1)}{6} \]

Combining the fractions gives:

\[ \frac{n(n+1)(2n+1 - 3)}{6} = \frac{n(n+1)(2n - 2)}{6} \]

### Step 5: Factor out the expression

Factoring out \(2\) from the numerator:

\[ \frac{n(n+1) \cdot 2(n - 1)}{6} = \frac{n(n+1)(n - 1)}{3} \]

### Final Result

Thus, the final result of the given expression is:

\[ \frac{n(n+1)(n-1)}{3} \]