- #1

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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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- Thread starter rought
- Start date

- #1

- 34

- 0

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

I have no idea on this one....

I do know that the sum formula is Sn=n(t1+tn)/2

- #2

HallsofIvy

Science Advisor

Homework Helper

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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?

- #3

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- 0

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?

- #4

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- 0

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

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