Question

Help with calculus!

Answer 1

Compute the sum and the limit of the sum as n approaches infinity .

Answer 2

idk

Answer 3

@Best_Mathematician can u help

Answer 4

sorry

Answer 5

its ok i tagged you by accident

Answer 6

its k

Answer 7

ok so what grade u in...just making sure how to explain it to u

Answer 8

i will message to u when u message me back and put a picture of u

Answer 9

@Best_Mathematician i'm in 12th grade

Answer 10

ohk would you mind waiting sometime as i myself is working on my question will get u in short timr

Answer 11

okay i'll wait

Answer 12

message me back

Answer 13

are you there? @Best_Mathematician

Answer 14

well @SithsAndGiggles got it...go ahead and explain her those stuff

Answer 15

Okay i did it before but i got it wrong now i see my mistake

Answer 16

Do you still need help figuring out the limit?

Answer 17

yea thats the part that i got stucked on

Answer 18

i got the two terms right just the limit part

Answer 19

okay

Answer 20

From there, it's a very simple limit if you just divide through by the highest power of \(n\). You should get \(\dfrac{3}{6}=\dfrac{1}{2}\). I'll. check my answer again to make sure.

Answer 21

ok

Answer 22

yea its correct it 1/2

Answer 23

Do you have the actual answer available? I've checked with WolframAlpha and I'm getting something completely different: http://www.wolframalpha.com/input/?i=Limit%5BSum%5B%281%2Fn%29%28%28i%2Fn%29%5E2%2B%282i%29%2Fn%29%2C%7Bi%2C1%2Cn%7D%5D%2Cn-%3EInfinity%5D

Answer 24

no thats the answer i got before..

Answer 25

Just a sec. I may have skipped a step in my work above...

Answer 26

Okay, I've figured out my mistake. I'm going to delete my earlier work.

Answer 27

\[\lim_{n\to\infty}\sum_{i=1}^n \frac{1}{n}\left[\left(\frac{i}{n}\right)^2+2\left(\frac{i}{n}\right)\right]\\
\lim_{n\to\infty}\sum_{i=1}^n \frac{1}{n}\left[\frac{1}{n^2}i^2+\frac{2}{n}i\right]\\
\lim_{n\to\infty}\sum_{i=1}^n \left[\frac{1}{n^3}i^2+\frac{2}{n^2}i\right]\\
\lim_{n\to\infty} \left[\sum_{i=1}^n\frac{1}{n^3}i^2+\sum_{i=1}^n\frac{2}{n^2}i\right]\\
\lim_{n\to\infty} \left[\frac{1}{n^3}\color{red}{\sum_{i=1}^ni^2}+\frac{2}{n^2}\color{blue}{\sum_{i=1}^ni}\right]\\
\lim_{n\to\infty} \left[\frac{1}{n^3}\color{red}{\frac{n(n+1)(2n+1)}{6}}+\frac{2}{n^2}\color{blue}{\frac{n(n+1)}{2}}\right]\\
\lim_{n\to\infty} \left[\frac{(n+1)(2n+1)}{6n^2}+\frac{n+1}{n}\right]\\
\lim_{n\to\infty} \left[\frac{(n+1)(2n+1)}{6n^2}+\frac{6n(n+1)}{6n^2}\right]\\
\lim_{n\to\infty} \frac{2n^2+3n+1+6n^2+6n}{6n^2}\\
\lim_{n\to\infty} \frac{8n^2+9n+1}{6n^2}=\frac{8}{6}=\frac{4}{3}\]

Answer 28

okay i see