Find the sum of the series 1-3+5-7+-11+ +1001

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Homework Help Overview

The problem involves finding the sum of a series that alternates signs among odd integers, specifically the series 1-3+5-7+...+1001. There is some confusion regarding the correct sequence of terms, particularly the presence of the number 11 and whether it should be replaced with 9.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the nature of the series and whether it can be treated as an arithmetic sequence. There are attempts to clarify the sequence and to identify how many pairs of terms contribute to the sum.

Discussion Status

Some participants have provided insights into the structure of the series and have attempted to derive the number of terms involved. There is an ongoing exploration of the correct interpretation of the series and how to approach the summation.

Contextual Notes

There is a noted discrepancy in the series terms, particularly regarding the transition from 7 to 11, which has led to questions about the intended sequence. Participants are also considering the implications of this on the total count of terms and pairs in the series.

rought
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Homework Statement



Find the sum of the series 1-3+5-7+-11+...+1001


Homework Equations



I have no idea on this one...

I do know that the sum formula is Sn=n(t1+tn)/2
 
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That formula only applies to arithmetic sequences and this is not an arithmetic sequence.

The first thing you will need to do is clarify the sum: 1-3+5-7+-11+...+1001. the first numbers, in absolute value, 1, 3, 5, and 7 differ by 2 but then there is the jump to 11, 4 larger than 7. Is it supposed to be 1-3+5-7+9-11+...+1001? If so then you can rewrite it 1+ (5-3)+ (9-7)+ (13- 11)+ ...+ (1001-999)= 1+ 2+ 2+ 2+ ...+ 2. Now, how many "2"s are there? How many pairs of odd numbers are there from 5 to 1001?
 
HallsofIvy said:
That formula only applies to arithmetic sequences and this is not an arithmetic sequence.

The first thing you will need to do is clarify the sum: 1-3+5-7+-11+...+1001. the first numbers, in absolute value, 1, 3, 5, and 7 differ by 2 but then there is the jump to 11, 4 larger than 7. Is it supposed to be 1-3+5-7+9-11+...+1001? If so then you can rewrite it 1+ (5-3)+ (9-7)+ (13- 11)+ ...+ (1001-999)= 1+ 2+ 2+ 2+ ...+ 2. Now, how many "2"s are there? How many pairs of odd numbers are there from 5 to 1001?


Yes sorry, there is supposed to be a 9 in there...

So there would be 999 "2"s ?

Would there be 200 pairs of odd numbers?
 
an=a+(n-1)d
1001=3+(n-1)2
1001-3=(n-1)2
998/2=n-1
499=n-1
499+1=n
n=500

500 odd integers from 3-1001

Therefore,
250 2s

250*2=500
500+1=501

Therefore,
the sum of the sequence=501 :wink:
 

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