PUMaC 2015 · 加试 · 第 1 题

PUMaC 2015 — Power Round — Problem 1

Your solutions are to be turned in when your team checks in on the morning of PUMaC or emailed to us at pumac@math.princeton.edu by 8AM Eastern Standard Time on the morning of PUMaC, November 21, 2015 with the subject line “PUMaC 2015 Power Round.” Please staple your solutions together, including the Power Round cover sheet (the last page of this document) as the first page. Each page should also have on it the team number (not team name) and problem number . Solutions to problems may span multiple pages, but staple them in continuing order of proof.

解析

This is a purely arithmetic fact. Most derivations all likely involve two instances of substitutions of s k 2 and a a = a a + a . One example is shown where the substituted parts are underlined: k k − 4 k − 3 k − 1 k − 2 2 2 a s = a s n − 1 n n n − 1 2 2 2 2 a a a − a a a = a a a − a a a n − 4 n +2 n − 3 n +1 n − 3 n +3 n − 2 n +2 n n n − 1 n − 1 2 2 2 2 a ( a a + a a ) = a ( a a + a a ) n +2 n − 4 n − 2 n − 3 n +3 n +1 n n − 1 n − 1 n 2 2 2 2 a ( a ( a a + a ) + a a ) = a ( a ( a a + a ) + a a ) n +2 n n − 1 n − 3 n − 2 n − 3 n − 1 n n +2 n +1 n − 2 n − 1 n +1 n 2 2 2 2 a a a a + a a a + a a a = a a a a + a a a + a a a n − 3 n − 1 n n +2 n n +2 n − 2 n +2 n − 3 n − 1 n n +2 n − 3 n +1 n − 3 n − 1 n − 2 n − 1 n n +1 2 2 a a ( a a + a ) = a a ( a a a ) n − 2 n +1 n − 2 n n − 3 n +1 n − 1 n +1 n − 1 n a a a a = a a a a . n − 3 n − 2 n +1 n +2 n − 3 n − 2 n +1 n +2