Library / Berkeley Math Circle Monthly Contest 7 / 2 of 15
Berkeley Math Circle Monthly Contest 7 Difficulty 4.3 AIME Prove it United States
Problem:f ( 1 ) = 1 f(1) = 1 f ( 1 ) = 1 f ( 1 ) + f ( 2 ) + ⋯ + f ( n ) = n 2 f ( n ) f(1) + f(2) + \cdots + f(n) = n^{2} f(n) f ( 1 ) + f ( 2 ) + ⋯ + f ( n ) = n 2 f ( n ) n ≥ 2 n \geq 2 n ≥ 2 f ( 2008 ) f(2008) f ( 2008 )
Solution Solution:n 2 f ( n ) − f ( n ) = f ( 1 ) + f ( 2 ) + ⋯ + f ( n − 1 ) = ( n − 1 ) 2 f ( n − 1 ) n^{2} f(n) - f(n) = f(1) + f(2) + \cdots + f(n-1) = (n-1)^{2} f(n-1) n 2 f ( n ) − f ( n ) = f ( 1 ) + f ( 2 ) + ⋯ + f ( n − 1 ) = ( n − 1 ) 2 f ( n − 1 ) f ( n ) = ( n − 1 ) 2 n 2 − 1 f ( n − 1 ) = n − 1 n + 1 f ( n − 1 ) f(n) = \frac{(n-1)^{2}}{n^{2} - 1} f(n-1) = \frac{n-1}{n+1} f(n-1) f ( n ) = n 2 − 1 ( n − 1 ) 2 f ( n − 1 ) = n + 1 n − 1 f ( n − 1 )
Thusf ( 2008 ) = 2007 2009 f ( 2007 ) = 2007 2009 ⋅ 2006 2008 f ( 2006 ) = 2007 2009 ⋅ 2006 2008 ⋅ 2005 2007 f ( 2005 ) = ⋯ = 2007 ! 2009 ⋅ 2008 ⋯ 4 ⋅ 3 f ( 1 ) = 2 2009 ⋅ 2008 \begin{aligned} f(2008) &= \frac{2007}{2009} f(2007) \\ &= \frac{2007}{2009} \cdot \frac{2006}{2008} f(2006) \\ &= \frac{2007}{2009} \cdot \frac{2006}{2008} \cdot \frac{2005}{2007} f(2005) \\ &= \cdots \\ &= \frac{2007!}{2009 \cdot 2008 \cdots 4 \cdot 3} f(1) \\ &= \frac{2}{2009 \cdot 2008} \end{aligned} f ( 2008 ) = 2009 2007 f ( 2007 ) = 2009 2007 ⋅ 2008 2006 f ( 2006 ) = 2009 2007 ⋅ 2008 2006 ⋅ 2007 2005 f ( 2005 ) = ⋯ = 2009 ⋅ 2008 ⋯ 4 ⋅ 3 2007 ! f ( 1 ) = 2009 ⋅ 2008 2
← Previous Browse the library Next →
Want a route through all this instead of an archive?
The track
puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.
Source: MathNet ,
licensed CC-BY-4.0 .
Statement reproduced verbatim; metadata (topic, difficulty) added by this project.