The value of `sum_(r = 16)^(30)(r + 2)(r - 3)` is equal to :

Options

A. 7770
B. 7785
C. 7775
D. 7780

Answer

To solve the problem, we need to evaluate the summation:

\[ \sum_{r=16}^{30} (r + 2)(r - 3) \]

### Step 1: Expand the expression
First, we expand the expression \((r + 2)(r - 3)\):

\[ (r + 2)(r - 3) = r^2 - 3r + 2r - 6 = r^2 - r - 6 \]

### Step 2: Rewrite the summation
Now, we can rewrite the summation as:

\[ \sum_{r=16}^{30} (r^2 - r - 6) \]

This can be separated into three different summations:

\[ \sum_{r=16}^{30} r^2 - \sum_{r=16}^{30} r - \sum_{r=16}^{30} 6 \]

### Step 3: Calculate each summation
1. **Calculate \(\sum_{r=16}^{30} r^2\)**:
We can use the formula for the sum of squares of the first \(n\) natural numbers:

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

Thus, we find:

\[ \sum_{r=1}^{30} r^2 - \sum_{r=1}^{15} r^2 \]

For \(n = 30\):

\[ \sum_{r=1}^{30} r^2 = \frac{30 \cdot 31 \cdot 61}{6} = 9455 \]

For \(n = 15\):

\[ \sum_{r=1}^{15} r^2 = \frac{15 \cdot 16 \cdot 31}{6} = 1240 \]

Therefore:

\[ \sum_{r=16}^{30} r^2 = 9455 - 1240 = 8215 \]

2. **Calculate \(\sum_{r=16}^{30} r\)**:
We can use the formula for the sum of the first \(n\) natural numbers:

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

Thus, we find:

\[ \sum_{r=1}^{30} r - \sum_{r=1}^{15} r \]

For \(n = 30\):

\[ \sum_{r=1}^{30} r = \frac{30 \cdot 31}{2} = 465 \]

For \(n = 15\):

\[ \sum_{r=1}^{15} r = \frac{15 \cdot 16}{2} = 120 \]

Therefore:

\[ \sum_{r=16}^{30} r = 465 - 120 = 345 \]

3. **Calculate \(\sum_{r=16}^{30} 6\)**:
Since 6 is a constant, we can multiply it by the number of terms in the summation:

The number of terms from 16 to 30 is \(30 - 16 + 1 = 15\).

Thus:

\[ \sum_{r=16}^{30} 6 = 6 \times 15 = 90 \]

### Step 4: Combine the results
Now we can combine all the results:

\[ \sum_{r=16}^{30} (r^2 - r - 6) = 8215 - 345 - 90 \]

Calculating this gives:

\[ 8215 - 345 = 7870 \]
\[ 7870 - 90 = 7780 \]

### Final Answer
Thus, the value of the summation is:

\[ \boxed{7780} \]