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!
 A: $$\lim_{n\rightarrow \infty}\sum_{k=1}^{n}[2+\frac{3}{n}k]^2\left(\frac{3}{n}\right)$$
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
$$3\int_0^1[2+3x]^2 dx = \color{red}{39}$$ 
A: Two approaches:


*

*Expand and take limit. The answer will be 39

*Use integration. This will be equal to $3\int_0^1(3x + 2)^2 dx = 39$
I will type out the steps after a while.
A: $$\sum_{k=1}^n\left(2+3\frac kn\right)^2\frac3n\xrightarrow[n\to\infty]{}3\int_0^1\left(2+3x\right)^2dx=\left.\frac13(2+3x)^3\right|_0^1=39$$
A: Before taking limits, your sum is
$$ \sum_{k=1}^n \left( \dfrac{12}{n} + \dfrac{36 k}{n^2} + \dfrac{27 k^2}{n^3} \right) $$
Using the formulas
$$ \eqalign{\sum_{k=1}^n 1 &= n\cr
            \sum_{k=1}^n k &= \dfrac{n(n+1)}{2}\cr
            \sum_{k=1}^n k^2 &= \dfrac{n(n+1)(2n+1)}{6}\cr}$$
this simplifies to 
$$ \dfrac{78 n^2 + 63 n + 9}{2 n^2} = \dfrac{78  + 63 n^{-1} + 9 n^{-2}}{2}$$
so the limit as $n \to \infty$ is $78/2 = 39$.
