PUMaC 2014 · 代数（B 组） · 第 5 题

PUMaC 2014 — Algebra (Division B) — Problem 5

[ 5 ] Given that a a − a + a − na = − n + 3 n − 1 and a = 1 , a = 3, find a . n n − 2 n n − 2 0 1 20 n − 1 (⌊ ⌋) (⌈ ⌉) n n

解析

[ 5 ] Given that a a − a + a − na = − n + 3 n − 1 and a = 1 , a = 3, find a . n n − 2 n n − 2 0 1 20 n − 1 Solution: 2 2 Factoring a bit, we see that a ( a − n ) − ( a − ( n − 1) )+( a − n ) = 0. Letting b = a − n , n − 2 n n n n n − 1 we have that, after some simplifying, that 2 b b − b + ( n − 1)( b − 2 b ) = 0 n − 2 n n n − 1 n − 1 n We have that b = 1 , b = 2, and also b = 4 so we guess b = 2 . By induction, we have 0 1 2 n n − 2 2 n − 2 n n n 20 2 b − 2 + ( n − 1)( b − 2 ) = 0 ⇒ b = 2 . Therefore a = 2 + n ⇒ a = 2 + 20 = n n n n 20 1048596 . (⌊ ⌋) (⌈ ⌉) n n