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3/n

okay, I'm understanding...

Are you still with me?

yes

((9i^2/n^2)+1)(3/n)

I hope you used \[[(\frac{ 3i }{ n })^2+1](\frac{ 3 }{ n })\] because i i mistypoed a 2

really makes you appreciate the fundamental theorem of calculus, doesn't it?

that's what I used... and no :( lol

no? compare with
\[F(x)=\frac{x^3}{3}+x\] and then \(F(3)-F(0)\)

Oh wait, nevermid, you were right. Now distribute. Haha.

(27i^2/n^3)+(3/n)

\[(9(n+1)(2n+1)/2n^2) + 3\]

Combined it all into one fraction*

umm... I have trouble finding the limit

unless a typed it wrong of course .... latex is nice but consumes alot of brain power

@amistre64 how did you do the "slash" across? :P

lol,
```
\cancel{arg}
```

and i put an ^0 on it to indicate it canceld

awe snaps!! this just got real!! Thanks :)

Do you have a latex site for these codes? I have one but it's so basic. Lol.

NEAT!