How to solve this: $\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left(\frac{2n+3k}{n} \right)^2$

Question

How to solve this:

$$\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left(\frac{2n+3k}{n} \right)^2$$

The answer is supposed to be 39.

My attempt:

$$\frac{3}{n}\sum\left(4+\frac{12}{n}k+\frac{9}{n^{2}}k^{2}\right)=\frac{3}{n}\left(4+\frac{12}{n}\sum k+\frac{9}{n^{2}}\sum k^{2}\right)=\frac{3}{n}\left(4+\frac{12}{n}\frac{n}{2}\left(n+1\right)+\frac{9}{n^{2}}\frac{n}{6}\left(n+1\right)\left(2n+1\right)\right)=3\left(6+3\right)=27\neq39$$

@ClaudeLeibovici, thanks! $\sum 4 = 4n => 3 (4 + 6 + 3) = 39$. — Yal dc, Jul 19 '14 at 08:53
You got it ! With respect to your comment to lab's answer, you made a mistake developing $(2+3x)^2$. And you are welcome. — Claude Leibovici, Jul 19 '14 at 08:56

score 5 · Accepted Answer · answered Jul 19 '14 at 08:48

5

$$\lim_{n\to\infty}\frac3n\sum_{k=1}^n\left(2+3\frac kn\right)^2$$

$$=3\int_0^1(2+3x)^2\ dx$$

$$\text{as }\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$

answered Jul 19 '14 at 08:48

lab bhattacharjee

274,582

$3\int (4+6x+9x^2) = 3 (4x+3x^2+3x^3) = 3 (4+3+3) = 30 \neq 39$ – Yal dc Jul 19 '14 at 08:51
@Yaldc, See also: http://math.stackexchange.com/questions/469885/the-limit-of-a-sum-sum-k-1n-fracnn2k2 – lab bhattacharjee Jul 19 '14 at 08:53
3

$$3\int (4+12x+9x^2) = 3 (4x+6x^2+3x^3) = 3 (4+6+3) = 39$$ – Claude Leibovici Jul 19 '14 at 08:54
@ClaudeLeibovici, There was a typo, it should be : $$3\int_0^1 (2+3x)^2\ dx=\frac{(2+3x)^3|_0^1}3=\frac{5^3-2^3}3 $$ – lab bhattacharjee Jul 19 '14 at 08:59

How to solve this: $\lim_{n \rightarrow \infty} \frac{3}{n} \sum_{k=1}^n \left(\frac{2n+3k}{n} \right)^2$

1 Answers1