12th APMO 2000

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Problem 1

Find a13/(1 - 3a1 + 3a12) + a23/(1 - 3a2 + 3a22) + ... + a1013/(1 - 3a101 + 3a1012), where an = n/101.

 

Solution

Answer: 51.

The nth term is an3/(1 - 3an + 3an2) = an3/( (1 - an)3 + an3) = n3/( (101 - n)3 + n3). Hence the sum of the nth and (101-n)th terms is 1. Thus the sum from n = 1 to 100 is 50. The last term is 1, so the total sum is 51.

 


 

12th APMO 2000

© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002