# Thread: Finding the sum of a series

1. ## Finding the sum of a series

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2. Originally Posted by SarahGr
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
1 - 3 + 5 - 7 + ... - 995 + 997 - 999 + 1001 = ?

sum, Gauss style ...

(1001 + 1) - (999 + 3) + (997 + 5) - (995 + 7) + ...

now, you have to determine if the individual terms all match up (yielding a sum of zero), or is there a term "left over" ?

3. Originally Posted by SarahGr
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
Is this supposed to be the alternating sign sum of the odd integers up to 1001?

If so write it as the difference of two arithmetic progressions:

(1 + 5 + 9 + ... + 1001) - (3 + 7 + 11 + ... + 999)

CB

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5. Originally Posted by SarahGr
Is there a formula that i can use to do those out? I'm not sure I understand what you mean
The sum of an arithmetic sequence is $S_n = \frac{n}{2}(2a+(n-1)d)$

where n is the number of terms, a is the first term and d is the common difference

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7. Originally Posted by SarahGr
ahh I'm so sorry guys the sequence is actually 1-3+5-7+7-11+...+1001

(I forgot the 7 =/)
This "+7" would be weird; I can't see the pattern.

For $S=1-3+5-7+\cdots+1001$, there is an easy way: group the terms 2 by 2, you get $S=(1-3)+(5-7)+(9-11)+\cdots+(997-999)+1001$. As you can see, all the parentheses equal -2, and there are ?? of them, so...

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