Find the sum when n=1000. Show all your steps. Use the properties of summation.

i=1 -->n ((4i/n^2)+(i^3/n))*(6/i)

Thank you very much....I hope I hear back asap!!!

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- Dec 5th 2010, 10:40 PMkparson4Summations
Find the sum when n=1000. Show all your steps. Use the properties of summation.

i=1 -->n ((4i/n^2)+(i^3/n))*(6/i)

Thank you very much....I hope I hear back asap!!! - Dec 6th 2010, 05:27 AMAckbeet
You're performing this sum, then?

$\displaystyle \displaystyle\sum_{i=1}^{n}\left[\left(\frac{4i}{n^{2}}+\frac{i^{3}}{n}\right)\frac {6}{i}\right]?$

If so, I would note two things:

1. You can simplify a bit before summing.

2. It's true from principles of induction that

$\displaystyle \displaystyle\sum_{i=1}^{n}i^{2}=\frac{n(n+1)(2n+1 )}{6}.$

So what does that tell you? - Dec 6th 2010, 05:59 AMkparson4Thank you
Thank you so much...you saved me!!!!

- Dec 6th 2010, 06:00 AMAckbeet
You're welcome. You have solved it, then?

- Dec 6th 2010, 06:05 AMkparson4
No I did not sadly, and it is due in one hour. Can you direct me anymore?

- Dec 6th 2010, 06:06 AMAckbeet
If it's "due", then it is for a grade, right? Forum policy is not knowingly to help with graded problems.

- Dec 6th 2010, 06:09 AMkparson4
"Due" was not the right word. Sorry this is the first time I have used this site. It is a practice problem given before our classes. My apologies, thanks for that information. Not for a grade.

- Dec 6th 2010, 07:22 AMAckbeet
Well, what simplifications do you get?