Sigma Notation Help?


by missadorkable
Tags: calculus, sigma
missadorkable
missadorkable is offline
#1
Mar27-11, 11:40 PM
P: 3
Express this in sigma notation?
3^3 - 3^4 + 3^5 - ... - 3^100

Evaluate these two sigmas?
n
∑ (i-2)^2
i =1

n
∑ (4-i^2)
i =1

I don't really understand sigma notation so I'm really interested in the process and explanation of how to do it. Any help would be greatly appreciated!
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Stephen Tashi
Stephen Tashi is offline
#2
Mar28-11, 12:02 AM
Sci Advisor
P: 3,175
Did your course materials cover how to add up the squares of consecutive integers?
Do they give a formula for
[tex] \sum_{i=1}^n i^2 [/tex]
missadorkable
missadorkable is offline
#3
Mar28-11, 12:09 AM
P: 3
yes.

n
∑ (i^2) = n(n+1)/2
i =1

How do I apply this? :S

Stephen Tashi
Stephen Tashi is offline
#4
Mar28-11, 12:20 AM
Sci Advisor
P: 3,175

Sigma Notation Help?


For example

[tex] \sum_{i=1}^n 3i^2 + 2i + 1 [/tex]

The summation can be distributed to obtain

[tex] = 3 \sum_{i=1}^n i^2 + 2 \sum_{i=1}^n i + \sum_{i=1}^n 1 [/tex]

[tex] = 3 ( n (n+1)/2) + 2 \sum_{i=1}^n i + n [/tex]

The materials probably give the the formula for [tex] \sum_{i=1}^n i [/tex] and you can use it also.

In the problems you asked about, you need to multiply out the expressions like [tex] (i-2)^2 [/tex] before you distribute the summation.
missadorkable
missadorkable is offline
#5
Mar28-11, 12:24 AM
P: 3
Alright, I think I understand what you're saying. What about the first question with expressing 3^3 - 3^4 + 3^5 - ... - 3^100 as sigma notation? Is there a formula that could be used to acquire the expression for the sigma notation?
Stephen Tashi
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#6
Mar28-11, 12:45 AM
Sci Advisor
P: 3,175
Quote Quote by missadorkable View Post
Is there a formula that could be used to acquire the expression for the sigma notation?
No, there isn't a mechanical procedure for determining the formula (the function of i) that would be used in the sigma notation to get that series. You have to do it by trial and error. The terms alternate, so it could involve a negative number raised to a power. You can throw in a factor of [tex] (-1)^i [/tex] or [tex] (-1)^{i+1} [/tex] in the formula if it is needed to make the signs come out correctly. The things that are increasing by 1 in each term should tell you that the formula involves [tex]i [/tex] as an exponent. It looks like the first exponent starts as 3 not as 1. If your materials want you to start all summations with [tex] i = 1 [/tex] then use the expression [tex] i+2 [/tex] in the exponent to make the first term have an exponent of 3.
Robert1986
Robert1986 is offline
#7
Mar28-11, 05:19 AM
P: 828
Quote Quote by missadorkable View Post
yes.

n
∑ (i^2) = n(n+1)/2
i =1

How do I apply this? :S
This is wrong. n(n+1)/2 is the sum of the first n integers.

To write the first in sigma notation, just look at the pattern. you've got a coefficient that is oscliating between -1 and 1, you have some odd numbers in the denominator. Just try to find a pattern.


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