# Simplfication of a sigma notation

AerospaceEng

## Homework Statement

1/1(4) + 1/4(7) + 1/7(10)+....+ 1/(3n-2)(3n+1)

and the sigma notation is pretty obvious

nothing really..

## The Attempt at a Solution

I can see an obvious pattern its n(n+3) but n cannot be 2,3,5,6 etc.. and the second digit in the first term becomes the first digit in the next term. I think its fairly simply but I've just been staring at it blankly and i think ive passed that point where im doing any useful thinking so an answer or even a hint would be great.

## Answers and Replies

Homework Helper
Hi AerospaceEng! Are you asking how to write it with a ∑ ?

If so, the answer's in the question: it's ∑ 1/(3n-2)(3n+1) (and btw the next step is probably to use partial fractions)

AerospaceEng
no lol but thank you, i definitely know how to write it in sigma but now that i have it in a "complex" sigma form I need to change it to a more simplified sigma notation so like i tried:

∑ 1/(n)(n+3) but that doesnt work for 2,3 and so on like i mentionned before

and then after wards i have to prove that my simplification works by the principle of mathematical induction. But i can do that part its just changing the sigma

Homework Helper
oh good! in that case, it is partial fractions …

ie, ∑ [ A/(3n-2) + B/(3n+1)) ] AerospaceEng
no, I don't think so. well not in my case anyways I haven't learnt that.
Thanks for trying, I'll post the answer once I figure it out to clarify what i needed.

Homework Helper
Well, try it anyway …

find numbers A and B such that A/(3n-2) + B/(3n+1) = 1/(3n-2)(3n+1)) (and you could also look at the examples at http://en.wikipedia.org/wiki/Partial_fractions#Examples")

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AerospaceEng
Okay so I have the answer now the sum of the series is equal to n/(3n+1) thats what i needed. but thanks anyways tim I really appreciate it