Heyy Forum Im having difficulty figuring out this question soo any help would be greatly appriciated Thanks so much
Last edited by jess0517; Jul 13th 2010 at 09:26 AM. Reason: SOLVED
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Did you try working out the product for different values of ?
Originally Posted by roninpro Did you try working out the product for different values of ? Yes i did but i still couldnt seem to figure it out
I computed the values for : Do you see a pattern, at least in the denominators?
Originally Posted by jess0517 Heyy Forum Im having difficulty figuring out this question soo any help would be greatly appriciated Thanks so much Do some particular cases: , , etc., and now separate in odd and even n's...can you see a pretty clear patern for each case? Tonio
Hello, jess0517! Find a prove a formula for: . I cranked out the first few partial products: . . It appears that: . I'll let you supply the proof.
I assume "find and prove" means that induction is involved. But anyway you could find the value of by direct manipulation of the terms of the product .
Notice that . Therefore Now start cancelling: each of the two factors in the numerators cancels with a factor in a neighbouring denominator. The only terms left uncancelled are at the start and at the end.
Originally Posted by Soroban Hello, jess0517! I cranked out the first few partial products: . . It appears that: . I'll let you supply the proof. So using induction would this be correct?.. im not to sure how to write it out formaly on this site but this is how i did it: [(k+1)+1]/2k =[k+2]/2k = 1/2+1/k = [k+2]/2k ? Obviously adding a formal base step, induction hypothesis and induction step
Thread closed due to this member deleting questions after getting help.
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