A rubber band is 1 meter long. An ant starts at one end, crawling at 1 millimeter per second. At the end of each second, the rubber band is instantaneously stretched by an additional meter. (So, at the end of the nth second, the rubber band becomes n+1 meters long.)

Does the ant ever reach the far end of the band? If so, when?

(In reply to The backward ant by TomM)

I found a pattern in your workings (else i wouldn't have seen anything at all) and you would have if you didn't simplify your workings. Appearantly, you could have done it forwards too.

At time T-3, the ant would have ((d(T-1)/T)+1)(T-2)/T-1)+1)(T-3)/(T-2). It is very hard to see it on this because it's flat but one can see that some of the numerators and denominators cancel out when the whole thing is simplified.

So the whole equation is:(leaving out some opening brackets because i don't know how many there are)
(d(T-1)/T)+1)(T-2)/T-1)+1)(T-3)/(T-2)+1)..... 4/3)+1)3/2)+1)2/1)+1= ?? (Don't know what to put in here because i can't find anything that won't contradict my next step)
Looking at the equation shows a way to get the ant moving forward.

(1(2/1)+1)3/2)+1)4/3)....+1)T/T-1)+d=1000+T1000
This simplifies to:
T+(T/2)+(T/3)+(T/4)+(T/5)+...+(T/(T-1))+d=1000+T1000
Factorise out T
T(1+(1/2)+(1/3)+(1/4)+(1/5)+...+(1/(T-1))+d=1000+T1000

I think there must be a flaw in either one or both my workings because there always is as all my other comments have shown. And i don't know how to solve this equation because I'm not sure if there is a method to add all the numbers up like geometric progressions...
If i am spouting babble that everyone knows, i'm sorry...

Posted by Aeternus on 2002-10-04 05:15:08