The value of ∑_{r=1}^{30} (r+2)(r-1) is equal to :

Let S = ∑{r=1}^{30} (r+2)(r-1) = ∑{r=1}^{30} (r^2 + r - 2)
Using the formulas:
∑{i=1}^{n} i^2 = n(n+1)(2n+1)/6
∑{i=1}^{n} i = n(n+1)/2
We have:
S = ∑{r=1}^{30} (r^2 + r - 2) = ∑{r=1}^{30} r^2 + ∑{r=1}^{30} r - ∑{r=1}^{30} 2
= [30(30+1)(60+1)/6] + [30(30+1)/2] - [2 * 30]
= [30(31)(61)/6] + [30(31)/2] - 60
= [9455] + [465] - 60
= 9860 - 60
= 9800
There seems to be a mistake in the question or the provided solution. Let's re-examine the expression:
∑{r=16}^{30} (r+2)(r-1) = ∑{r=16}^{30} (r^2 + r - 2)
= ∑{r=1}^{30} (r^2 + r - 2) - ∑{r=1}^{15} (r^2 + r - 2)
= [∑{r=1}^{30} r^2 + ∑{r=1}^{30} r - 2(30)] - [∑{r=1}^{15} r^2 + ∑{r=1}^{15} r - 2(15)]
= [30(31)(61)/6 + 30(31)/2 - 60] - [15(16)(31)/6 + 15(16)/2 - 30]
= [9455 + 465 - 60] - [1240 + 120 - 30]
= 9860 - 1330
= 8530
The given options are incorrect. There might be an error in either the question or the options.