9367 Q38

Forum for the GRE subject test in mathematics.
yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

9367 Q38

Postby yoyobarn » 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?

echo
Posts: 18
Joined: Fri Dec 16, 2011 7:41 pm

Re: 9367 Q38

Postby echo » Fri Dec 16, 2011 7:49 pm

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.

yoyobarn
Posts: 80
Joined: Sun Dec 19, 2010 7:01 am

Re: 9367 Q38

Postby yoyobarn » Sat Dec 17, 2011 2:41 am

Thanks!

That'll save many minutes of time!




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