Calculating the Sum of a Series with Sigma Notation

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Homework Help Overview

The discussion revolves around calculating the sum of a series using sigma notation, specifically focusing on the interpretation of terms and limits in the context of a summation from N=3 to N=7.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the implications of having a constant term in the sigma notation and question how it interacts with the variable limits. There are attempts to clarify whether the sum should simply yield the constant or if it should include additional terms based on the limits provided.

Discussion Status

Participants are actively engaging with each other's reasoning, with some providing alternative perspectives on the interpretation of the sigma notation. There is a recognition of the need to account for all terms in the summation, and guidance has been offered regarding the correct number of terms involved.

Contextual Notes

Some participants express uncertainty about the role of the variable N in the summation when a constant is present, leading to discussions about the nature of the terms being summed and the overall structure of the series.

xLaser
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7
----
\
/ 1 <- *note: the 1 should be more to the right.
----
N=3


Ok, this might be dumb :blushing: , but it think the answer for this is 1 but I'm just making sure.

cuz i know that for example if it is

7
----
\
/ N <- *note: the N should be more to the right.
----
N=3

the answer would be 3+4+5+6+7

but if it is just a 1 there, then is it just 1? i mean the N doesn't really matter because N isn't present as a variable on the right side.

Thx guy,

Mike.
 
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Actually, the answer is not 1.

Here, try this : find

[tex]\sum_{N=3}^7 (N+1)[/tex]
and use the fact that this should be equal to [tex]\sum_{N=3}^7 N + \sum_{N=3}^7 1[/tex].

From this you can find the last term, which is the one you want.
 
Last edited:
wouldn't that be :

(3+4+5+6+7)+(1) ??

because

7
----
\
/ 1 <- *note: the 1 should be more to the right.
----
N=3

is = to 1 no matter what since there is no variable there?@@#%?!
 
When N=3, what is (N+1) = ? And what about when N= 4 ? ...
 
ok i get your question above, but how about if it is
7
----
\
/ 1 <- *note: the 1 should be more to the right.
----
N=3

by itself? this sigma by itself would just mean 1 rite?
 
and it doesn't have to be limited to 1 only, what about if it is 2 there, or 3, or 4..? There is no variable N there so the N shouldn't really do anything... for any value of N the value would still be the number on the right?! doesn't that make sense?
 
ooooooooooooooooooooooooooooooooooooooook i just figured it out.

for

7
----
\
/ 1 <- *note: the 1 should be more to the right.
----
N=3

it actually means t1 + t2 + t3 + t4

where t1 = 1, t2 =1 and so on,

therefore, the answer should be

1+1+1+1 = 4 !

is that rite Gokul43201
 
It's not 4 either, but you're closer.

Let [tex]t_i=1[/tex] for all values of i. Your sum is

[tex]\sum_{i=3}^{7}t_i=t_3+t_4+t_5+t_6+t_7[/tex]

Note that you had missed a term when you rewrote it as t1+t2+t3+t4. There are actually 5, not 4, terms in this sum. If you are going to change the index variables starting point, make sure you adjust the end carefully as well.
 
aah ok thanks a lot m8.
 

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