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

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The series 1-3+5-7+...+1001 was clarified to include 9, leading to the correct sequence of odd numbers. The discussion involved determining the number of pairs of odd integers from 3 to 1001, concluding there are 500 odd integers. This results in 250 pairs contributing a sum of 2 each. Adding the initial term 1 to the total gives a final sum of 501 for the series. The solution emphasizes the importance of correctly identifying the sequence and counting pairs accurately.
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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:
 
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
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