Question

given that \[B_{n} = \sum_{i=0}^{n}(2/n)(4 - ((4i^2) / (n^2)))\] is the Riemann sum of an integrable function f(r) on an interval I. Find f(r) and I then evaluate \[\lim_{n \rightarrow \infty}B_{n}\] by integrating f(r) on I.

Answer 1

another cool question :)

Answer 2

this one?

\[\sum _{i=0}^n \left(\frac{2}{n}\left(\frac{4-4i^2}{n^2}\right)\right)\]

Answer 3

nono the 4 should be another rational k?

Answer 4

its 4 - (b / a) (prototype)

Answer 5

\[B_n = \sum_{1=0}^{n} \frac{2}{n}* (4 - \frac{4*i^2}{n^2})\]Right?

Answer 6

yes its right :D

Answer 7

Now, what am I supposed to do here

Answer 8

u should find f(r) and evaluate \[\lim_{n \rightarrow \infty}B _{n}\]

Answer 9

Should I integrate the whole function \[B_n = \sum_{i=0}^{n} \frac{2}{n}* (4 - \frac{4*i^2}{n^2})\] from 0 to n ...to find \(f(r)\)

Answer 10

hmm i don't know how can i solve this thats why i asked :D

Answer 11

Maybe we need integrating it, as for evaluating function \(n \rightarrow \infty\) we need to have some value for \(i\) too I mean we need some thing to get in place of \(i\)

Answer 12

ah yea we need

Answer 13

or not o.O

Answer 14

Ah I have a method ...

\[B_n = \sum_{i=0}^{n} \frac{8}{n} - \frac{8*i^2 }{n^3}\]
Now sum square of n natural numbers is \(\frac{n(n+1)(2n+1)}{6}\)
So I think the \(B_n\) thing should become
\[B_n = \frac{8}{n}*n - \frac{8}{n}*\frac{n(n+1)(2n+1)}{6}\]

Answer 15

hmm

Answer 16

ah yea

Answer 17

In this riemann sum, the width of each rectangle is \[\frac{2}{n}\] if 2 is the interval being evaluated. This would make the height of each rectangle equal\[f(i) = \left(2-\frac{2i}{n}\right)\left(2+\frac{2i}{n}\right)\]
The actual x value of for each i value is related as
\[x = i \times \frac{2}{n}\] so we can substitute that in the previous equation to find the original function being integrated:
\[f(x) = (2-x)(2+x)\]
You can then integrate it from 0 to 2\[\int\limits_{0}^{2}(4-x^{2})dx = 4x-\frac{x^{3}}{3} +C = 5.333... - 0 = 5.333... \]
I guess this is right, but I'm not sure...

Answer 18

\[B_n = \frac{8}{n}*n - \frac{8}{n^3}*\frac{n(n+1)(2n+1)}{6}\]

Answer 19

I didn't included powers typo

Answer 20

\[B_n = 8 - 8*\frac{2n^2 + 3n +1 }{6n^2 }\]

Answer 21

\[\lim_{n \rightarrow \infty} 8 - \frac{8}{6}*(2 + \frac{3}{n} +\frac{1}{n^2})\]

Answer 22

\[\frac{1}{n} \rightarrow 0\]

Answer 23

\[8 -\frac{8*2}{6}\]

Answer 24

yeah I get the same thing as alexray

Answer 25

5.33333333333333333333333333333333.........................

Answer 26

thanks anyway :D
i have like 200-300 questions like that lol

Answer 27

More 3s!!

Answer 28

lol @ alexray19

Answer 29

lol

Answer 30

Please do post more I want to solve them too

Answer 31

kk give me some time, i post one more .D :D

Answer 32

:D