
#1
Mar2711, 11:40 PM

P: 3

Express this in sigma notation?
3^3  3^4 + 3^5  ...  3^100 Evaluate these two sigmas? n ∑ (i2)^2 i =1 n ∑ (4i^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
Mar2811, 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] 



#3
Mar2811, 12:09 AM

P: 3

yes.
n ∑ (i^2) = n(n+1)/2 i =1 How do I apply this? :S 



#4
Mar2811, 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] (i2)^2 [/tex] before you distribute the summation. 



#5
Mar2811, 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?




#6
Mar2811, 12:45 AM

Sci Advisor
P: 3,175





#7
Mar2811, 05:19 AM

P: 828

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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