9367 Q38
Posted: Fri Dec 16, 2011 9:13 am
$$\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^n{[(\frac{3i}{n})^2-(\frac{3i}{n})]}=$$
What would be the shortest way to do this?
My long way is to simplify the summation:
$$\sum_{i=1}^n{\frac{9}{n^2}i^2-\frac{3}{n}i}=\frac{9}{n^2}\frac{1}{6}n(n+1)(2n+1)-\frac{3}{n}\frac{n^2+n}{2}=\frac{3(n+1)(2n+1)}{2n}-\frac{3(n+1)}{2}$$
Multiplying by 3/n gives
$$\frac{9(n+1)(2n+1)}{2n^2}-\frac{9(n+1)}{2n}=\frac{9(n+1)(2n+1)-9n(n+1)}{2n^2}$$
Considering the highest power (n^2), the term tends to (18-9)/2=9/2
(Option D)
Is there a shortcut?
What would be the shortest way to do this?
My long way is to simplify the summation:
$$\sum_{i=1}^n{\frac{9}{n^2}i^2-\frac{3}{n}i}=\frac{9}{n^2}\frac{1}{6}n(n+1)(2n+1)-\frac{3}{n}\frac{n^2+n}{2}=\frac{3(n+1)(2n+1)}{2n}-\frac{3(n+1)}{2}$$
Multiplying by 3/n gives
$$\frac{9(n+1)(2n+1)}{2n^2}-\frac{9(n+1)}{2n}=\frac{9(n+1)(2n+1)-9n(n+1)}{2n^2}$$
Considering the highest power (n^2), the term tends to (18-9)/2=9/2
(Option D)
Is there a shortcut?