- #1
mesa
Gold Member
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- 38
I have a series that takes steps of '2' which requires an operation starting from n=1 to do the following,
@n=1 (n-1)!
@n=3 (n-3)!-(n-1)!
@n=5 (n-5)!-(n-3)!+(n-1)!
@n=7 (n-7)!-(n-5)!+(n-3)!-(n-1)!
etc. etc.
Any ideas?
*EDIT*
Come to think of it,
This problem would probably be easier to solve if we split this thing up into two infinite series where one only has positive terms and the other all the negatives and then put them back together after the fact. I tried this with something similar not too long ago and it ended with a solution.
@n=1 (n-1)!
@n=3 (n-3)!-(n-1)!
@n=5 (n-5)!-(n-3)!+(n-1)!
@n=7 (n-7)!-(n-5)!+(n-3)!-(n-1)!
etc. etc.
Any ideas?
*EDIT*
Come to think of it,
This problem would probably be easier to solve if we split this thing up into two infinite series where one only has positive terms and the other all the negatives and then put them back together after the fact. I tried this with something similar not too long ago and it ended with a solution.
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