$$\lim_{n\to\infty} \frac{3}{n}\sum_{i=1}^n{[(\frac{3i}{n})^2-(\frac{3i}{n})]}=$$

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?

## 9367 Q38

### Re: 9367 Q38

You've done the problem the way calculus students hate to do it: solving the Riemann sum limit.

The quick way to do it is to realize that the limit describes an integral as a limit of Riemann sums. Specifically:

$$\int_0^3 (x^2-x) dx = 9/2$$

The 3/n tells you that the width of each of the rectangles and the inside of the summation shows the sampling points of the function.

The quick way to do it is to realize that the limit describes an integral as a limit of Riemann sums. Specifically:

$$\int_0^3 (x^2-x) dx = 9/2$$

The 3/n tells you that the width of each of the rectangles and the inside of the summation shows the sampling points of the function.

### Re: 9367 Q38

Thanks!

That'll save many minutes of time!

That'll save many minutes of time!