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anonymous
 3 years ago
how do i find the limit as an goes to infinity of the following? (the equation's in the comments)
anonymous
 3 years ago
how do i find the limit as an goes to infinity of the following? (the equation's in the comments)

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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 4 }{ n }\sum_{a=1}^{n}((\frac{ 4a }{ n })^{2} 4(\frac{ 4a }{ n }))\] This is the equation in question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 4 }{ n }\sum_{a=1}^{n}((\frac{ 4a }{ n })^{2})  \frac{ 4 }{ n }\sum_{a=1}^{n}(4(\frac{ 4a }{ n}))\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we're not sure where to go from here

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0looks like you are trying to compute some sort or reimann sum

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0exactly. a right riemann sum of x^2 4x to be exact.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0don't let the \(n\) through you off in the summation you are summing over \(a\) (never seen an \(a\) used like that before, but no matter) pull it right out front of the sum then use the formula for \[\sum_{a=1}^na^2\] and also \[\sum_{a=1}^na\]

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.0After a little algebra. \(\dfrac{16}{n^{2}}\left[\dfrac{1}{n}\sum\limits_{a=1}^{n}a^{2} + \sum\limits_{a=1}^{n}a\right]\)

primeralph
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 Got this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ 4 }{ n }\sum_{a=1}^{n}((\frac{ 4a }{ n })^{2})\] \[=\frac{64}{n^3}\sum_{a=1}^na^2\] for example @primeralph maybe, i am tired and all this latex is wearing me out

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@tkhunny i think the \(4\) is being squared as well

tkhunny
 3 years ago
Best ResponseYou've already chosen the best response.0Whoops. Make that 64, instead of 16.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0not really sure the sum is correct, but assuming it is for the moment \[\frac{64}{n^3}\sum_{k=1}^nk^2=\frac{64}{n^3}\frac{n(n+1)(2n+1)}{6}\] for the first one

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Wait but how does this help us get the limit as n > infinity?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0taking the limit as \(n\to \infty\) is very easy because the numerator is a polynomial of degree 3 will leading coefficient 128 and the denominator is a polynomial of degree 3 with leading coefficent 6

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0as you remember from some pre calc course about "horizontal asymptotes" that makes the limit \(\frac{128}{6}\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh oops duh that makes sense.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0easy if you think about it that way instead of the way they do it in the text by dividing top and bottom by \(n^3\) and all that other nonsense

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks so much!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@satellite73 where did 128 come from?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry probably a dumb question but...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oh waitn.....nevermind
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