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Question about Sigma and Sum Notation

  • Thread starter Bucs44
  • Start date
57
0
Here is the problem I'm at currently. I'm not sure that I'm on the right track or not. Also I'm not sure what numbers I should be plugging into the equation. I think it would be 2 through 6 but...?

The sum of elements in the set {ti | i = 3 } #7 on top of sigma notation

tn = 2n - 1, n greater than or equal to 1

Here is my calculation so far:

i1 + i2 + i3 + i4 + i5 + i6 + i7 = (1-1)+(2-1)+(3-1)+(4-1)+(5-1)+(6-1)+(7-1)

Where do I go next?
 

Answers and Replies

112
0
just use the formula sigma(i, i from 1 to n) = n(n+1)/2.

In this question, there are only 7 terms, so you can also add up by hand.
 
57
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just use the formula sigma(i, i from 1 to n) = n(n+1)/2.

In this question, there are only 7 terms, so you can also add up by hand.
I don't understand that - why would I be adding if ti = 2n - 1?
 
cristo
Staff Emeritus
Science Advisor
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I don't really understand your question. Is it calculate [tex]\sum_{n=1}^7t_n[/tex], where tn=2n-1?

If not, please could you quote the exact question as written. (NB, click on the formula to get the code for the LaTex equation)
 
57
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Yes sorry - I'm not able to write it exactly - don't have math software - but how you have shown it is correct
 
cristo
Staff Emeritus
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Yes sorry - I'm not able to write it exactly - don't have math software -
Note that the software is preloaded into the forum, and so anyone can write in LaTex. See here for a tutorial.

but how you have shown it is correct
If my sum above is correct, then you simply sum over n in the range 1 to 7. So, [tex]\sum_{n=1}^7(2n-1)= (2*1-1)+(2*2-1)+ \cdots[/tex] Do you see where I'm going here? Just plug in the remaining values of n, and then simplify the sum (to obtain a number) which will be the answer.
 
57
0
Note that the software is preloaded into the forum, and so anyone can write in LaTex. See here for a tutorial.



If my sum above is correct, then you simply sum over n in the range 1 to 7. So, [tex]\sum_{n=1}^7(2n-1)= (2*1-1)+(2*2-1)+ \cdots[/tex] Do you see where I'm going here? Just plug in the remaining values of n, and then simplify the sum (to obtain a number) which will be the answer.
Okay - so I do that up to 7 and then add them together or subtract?

I'd get 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49
 
57
0
Correct me if I'm wrong, but in order to obtain the product, I simply multiply those numbers for my total?
 
cristo
Staff Emeritus
Science Advisor
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Bucs44 said:
Okay - so I do that up to 7 and then add them together or subtract?

I'd get 1 + 3 + 5 + 7 + 9 + 11 + 13 = 49
Thats right, you add the terms.

Correct me if I'm wrong, but in order to obtain the product, I simply multiply those numbers for my total?
Is this a different question now? Do you want to find [tex]\Pi_{n=1}^7(2n-1)[/tex]? If so, yes, you would multiply the terms.
 

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