Question

Summation, please check my work!

Answer 1

Original Question

Answer 2

I suppose I made a mistake somewhere along there in the algebra D:

Answer 3

I see one error so far (see attached). I'm still looking through the pages though

Answer 4

oh thanks!

Answer 5

I am using right end point :)

Answer 6

ok so yeah you start at i = 1

Answer 7

http://www.wolframalpha.com/input/?i=expand++%28%281%2B2i%2Fn%29%5E3-%281%2B2i%2Fn%29%5E2%2B2%281%2B2i%2Fn%29-1%29&a=i_Variable

Answer 8

Yay! so I guess the first part is right :D Then it's plugging it into the limit and solving.

Answer 9

yes, everything looks good till that

Answer 10

but im pretty sure there is some arithmetic mistake somewhere in tha later pages

Answer 11

http://www.wolframalpha.com/input/?i=%5Csum%5Climits_%7Bi%3D1%7D%5E%7Bn%7D+%28%281%2B2i%2Fn%29%5E3-%281%2B2i%2Fn%29%5E2%2B2%281%2B2i%2Fn%29-1%29*2%2Fn

Answer 12

mm yeah. My professor said the answer should come out to = 52/3
so all the n's should cancel out as well.

Answer 13

another error

Answer 14

mm why? it was 4 previously o.o

Answer 15

ooh
so really that's not even a fraction anymore

Answer 16

because you went from 8n^4 to just 2n up top

you divided by 4n^3

same for the other terms

Answer 17

so it should just be 2n+4+2n^-1

Answer 18

yeah or \[\Large 2n+4+\frac{2}{n}\] for that bit

Answer 19

Excellent. So now the whole next page is wrong sigh xD

Answer 20

Actually on the next page i can keep the entire left of the + and then on the right side just erase the 4*6 underneath, yeah? and extend the line from the other part

Answer 21

its just an arithmetic error... i would give 9/10
your overall work for the riemann sum looks really good

Answer 22

must..get...perfect...score D: xD

Answer 23

so now on the first line of the second page (3rd picture of work)
= [24n+18+16n+8+16+8n^-1+12n+24+12]/6n

Answer 24

yeah the prof integrated in class today and the answer comes to 52/3

Answer 25

why not combine the 2n+4 with the stuff more towards the left?

(n+3n+3) + (2n+4)
(n+3n+2n) + (3+4)
6n+7

that should make things a bit simpler in my opinion

Answer 26

oh before multiplying by 6?

Answer 27

at the end of the second picture, yeah?

Answer 28

\[(6n+7+3+\frac{ 2 }{ n }+\frac{ 16n+32 }{ 6n})(2/n)\]

Answer 29

Here's one way to go about it

\[\large \left(6n+7+\frac{16n+8+16+8n^{-1}}{6}+\frac{2}{n}\right)*\left(\frac{2}{n}\right)\]

\[\large \left(6n+7+\frac{16n+24+8n^{-1}}{6}+\frac{2}{n}\right)*\left(\frac{2}{n}\right)\]

\[\large \left(6n+7+\frac{16n+24+8n^{-1}}{6}+\frac{2}{n}\right)*\color{red}{\left(\frac{2}{n}\right)}\]

\[\large 6n\color{red}{*\left(\frac{2}{n}\right)}+7\color{red}{*\left(\frac{2}{n}\right)}+\frac{16n+24+8n^{-1}}{6}\color{red}{*\left(\frac{2}{n}\right)}+\frac{2}{n}\color{red}{*\left(\frac{2}{n}\right)}\]

do you see what to do next?

Answer 30

em do we just multiply stuff out?

Answer 31

yes

Answer 32

so multiply the 2/n into everything and then put it all over 6?

Answer 33

yes that should work

Answer 34

ok just a moment

Answer 35

so my common denominator will be 6n^2?

Answer 36

yep

Answer 37

\[\frac{ 72n^2+84n+32n^2+48n+16+24 }{ 6n^2 }\]

Answer 38

\[\Large \frac{ 72n^2+84n+32n^2+48n+16+24 }{ 6n^2 }\]

is correct. Now combine like terms

Answer 39

[104n^2+132n+16]/6n^2

Answer 40

you forgot the +24

Answer 41

thanks! xD so that 16 becomes a 40

Answer 42

yes

Answer 43

Divide everything by 2n?

Answer 44

or maybe just 2

Answer 45

[52n^2+66n+20]/2n^2

Answer 46

why not break up the fraction?
\[\Large \frac{104n^2+132n+40}{6n^2} =\frac{104n^2}{6n^2} +\frac{132n}{6n^2} +\frac{40}{6n^2}\]

Answer 47

mm and then do what? we would get a weird number for the 40 one

Answer 48

simplify each fraction on the RHS

then you can take the limit as n --> infinity

Answer 49

oooo yes!!

Answer 50

so at the end should i write out the original problem = 53/2 ? just for closure

Answer 51

sure, that wraps things up nicely

Answer 52

thank you so much!

Answer 53

you're welcome