Sigma Notation Help?

  1. 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!
     
  2. jcsd
  3. Stephen Tashi

    Stephen Tashi 4,413
    Science Advisor
    2014 Award

    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]
     
  4. yes.

    n
    ∑ (i^2) = n(n+1)/2
    i =1

    How do I apply this? :S
     
  5. Stephen Tashi

    Stephen Tashi 4,413
    Science Advisor
    2014 Award

    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.
     
  6. 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?
     
  7. Stephen Tashi

    Stephen Tashi 4,413
    Science Advisor
    2014 Award

    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.
     
  8. 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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