Question

Find lim x--> infinity of (3/n)((3i+3)/n)^2

Answer 1

\[\lim_{x \rightarrow \infty} \sum_{i = 1}^{n}(\frac{ 3 }{ n })(\frac{ 3i+3 }{ n })^2\]

Answer 2

sorry forgot the summation in the question

Answer 3

Do you mean, as n goes to infinity?

Answer 4

There is no 'x' in the expression?

Answer 5

oh yeah sorry

Answer 6

I would start by expanding the expression. You can pull out any constants, as well as \(n\) from the summations.

Answer 7

Then recall that \[
\sum_{i=1}^n i = n(n+1)/2 \\
\sum_{i=1}i^2=n(n+1)(2n+1)/6
\]

Answer 8

@jennychan12 get it?

Answer 9

i'm working on it...

Answer 10

i'm expanding the inside for now. i didn't write lim n--> infinity or summation (i = 1 to n)
\[\frac{ 3 }{ n }(\frac{ 3i+3 }{ n })^2 = \frac{ 3 }{ n }(\frac{ 9i+18i+9 }{ n^2 })\]
\[\frac{ 27i+54i+9 }{ n^3 } = 27(\frac{ (\frac{ n(n+1) }{ 2 }) }{ n^3 }) + 54(\frac{ \frac{ n(n+1)(2n+1) }{ 6 } }{ n^3 })+9(\frac{ n }{ n^3 })\]
correct? or did i do something wrong?

Answer 11

whoops the last term should be 27(...)

Answer 12

If you type \left( \right) it will size the parenthesis nicely usually.

Answer 13

Now you just need to simplify the expression a bit more, so you can take the limit.

Answer 14

i did and i got the answer to be 54

Answer 15

The problem is that I don't understand because you should have \(i^2\) and I don't see it anywhere.

Answer 16

oh sorry that should be the first term.
oops. i plugged it in wrong.
but i did it right on my paper.
and i got 54..

Answer 17

but the answer is 9 so i did something wrong...

Answer 18

Try again. I can tell you where you went wrong if I can see your steps.

Answer 19

yeah i got the right answer. i forgot to divide 54 by 6 :/