Calculating the Sum of a Series with Sigma Notation

AI Thread Summary
The discussion revolves around understanding sigma notation and calculating the sum of a series. Initially, there is confusion about the sum when the upper limit is 7 and the lower limit is 3, particularly with the presence of a constant term. It is clarified that the sum should include all terms from the lower to the upper limit, leading to the conclusion that the sum equals 5, not 1 or 4. The importance of correctly identifying the number of terms in the series is emphasized, as well as the need to adjust indices properly when changing variables. Ultimately, the correct approach to sigma notation is reinforced through collaborative problem-solving.
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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

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

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 t_i=1 for all values of i. Your sum is

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

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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