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

[rought]

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

 

[HallsofIvy]

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?

 

[rought]

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?

 

[serectrus]

a_n=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