UW Challenge June 10
June 17th, 2008 by erb
This weeks problem: Determine A2008 if A0 = A and
An+1 = An / (1 + n An), for n = 0, 1, 2, …
Solution: A2008 = A / (1 + 2 015 028 A )
See my full solution PDF (With Mathematica and by-hand solutions).
One Response to “UW Challenge June 10”
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Anonymous
wrote on 06/21/08 at 9:28 pm :
Nice work! It is actually quite an easy problem: if you let
c[n] = 1 / A[n]
the given equation becomes
c[n+1] = c[n] + n for n = 0, 1, …
so c[2008] = c[0] + (1 + 2 + … + 2007).
Now, since (every freshman knows that)
1 + 2 + … + n = n (n + 1) / 2
we have
A[2008] = 1 / ( 1 / A + 2007 * 2008/2 ).