Calculus: How do you solve this double sequence as its limit approaches infinity? How do you solve the series below?
$$ \lim_{n\rightarrow \infty}\space\sum_{k=1}^{n}\space(2+\frac{3}{n}k)^2\space(\frac{3}{n}) $$
This question is a multiple choice question, and the choices are:
(A) 0
(B) 1
(C) 4
(D) 39
(E) 125
I had initially thought this was an infinite series question, but Thomas Andrews pointed out that this is a double sequence instead. I do not think double sequences are on the AP Calculus BC curriculum but I would also like to know how to solve them as well.
Thank you!

$$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}[2+\frac{3}{n}k]^2\left(\frac{3}{n}\right)$$
We first remember that
$$b\int_0^1 f(x) \, dx = \lim_{n \to \infty} \frac{b}{n}\sum_{i=1}^{\infty} f\left(\frac{i}{n}\right)$$
If we adjust the sum we get
$$\frac{3}{n}\lim_{n\rightarrow \infty}\sum_{i=1}^{n}\left[2+3\frac{i}{n}\right]^2$$
And we now transform this to
$$3\int_0^1[2+3x]^2 dx = \color{red}{39}$$ 
