The value of $$\sum_{r=16}^{30}(r+2)(r-3)$$ is equal to:

We have to evaluate the finite sum

$$\sum_{r=16}^{30}(r+2)(r-3).$$

First, expand the product inside the summation. Using the algebraic identity $$(a+b)(a+c)=a^2+(b+c)a+bc,$$ we set $a=r,\;b=2,\;c=-3$$ and obtain

$$\;(r+2)(r-3)=r^2+(2-3)r+2(-3)=r^2-r-6.$$

So the given sum becomes

$$\sum_{r=16}^{30}(r+2)(r-3)=\sum_{r=16}^{30}\bigl(r^2-r-6\bigr).$$

By the distributive property of summations, split this into three separate sums:

$$\sum_{r=16}^{30}r^2-\sum_{r=16}^{30}r-\sum_{r=16}^{30}6.$$

Let us evaluate each part one by one.

1. Sum of squares: We use the standard formula for the sum of the first $$n$$ squares,

$$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}.$$

To find $$\sum_{r=16}^{30}r^2$$, calculate $$\sum_{k=1}^{30}k^2$$ and subtract $$\sum_{k=1}^{15}k^2$$ because the terms from 1 to 15 are not required.

For $$n=30$$:

$$\sum_{k=1}^{30}k^2=\frac{30\,(30+1)\,(2\cdot30+1)}{6} =\frac{30\cdot31\cdot61}{6}.$$

Compute step by step: $$30\cdot31=930,\;930\cdot61=56\,730,$$ and finally $$\frac{56\,730}{6}=9\,455.$$

For $$n=15$$:

$$\sum_{k=1}^{15}k^2=\frac{15\,(15+1)\,(2\cdot15+1)}{6} =\frac{15\cdot16\cdot31}{6}.$$

Now evaluate: $$15\cdot16=240,\;240\cdot31=7\,440,$$ and $$\frac{7\,440}{6}=1\,240.$$

Therefore

$$\sum_{r=16}^{30}r^2=9\,455-1\,240=8\,215.$$

2. Sum of first powers: The sum of an arithmetic sequence from $$a$$ to $$b$$ with $$n$$ terms is $$\dfrac{n}{2}(a+b).$$ Here $$a=16,\;b=30,$$ and the number of terms is $$n=30-16+1=15.$$ Hence

$$\sum_{r=16}^{30}r=\frac{15}{2}(16+30)=\frac{15}{2}\cdot46=15\cdot23=345.$$

3. Sum of the constant $$-6$$: A constant summed $$n$$ times is simply the constant multiplied by $$n$$. As there are 15 terms,

$$\sum_{r=16}^{30}6 = 6\times15 = 90.$$

Because the constant in our expression is $$-6$$, we actually have $$-\sum6=-90.$$

Combine all the parts:

$$\sum_{r=16}^{30}(r^2-r-6)=\bigl(8\,215\bigr)-\bigl(345\bigr)-\bigl(90\bigr).$$

First subtract 345:

$$8\,215-345=7\,870.$$

Now subtract 90:

$$7\,870-90=7\,780.$$

Therefore,

$$\sum_{r=16}^{30}(r+2)(r-3)=7\,780.$$

Among the given options, 7 780 corresponds to Option C.

Hence, the correct answer is Option C.