HMMT 十一月 2018 · 冲刺赛 · 第 17 题

HMMT November 2018 — Guts Round — Problem 17

[ 10 ] Pascal has a triangle. In the n th row, there are n + 1 numbers a , a , a , . . . , a where n, 0 n, 1 n, 2 n,n a = a = 1. For all 1 ≤ k ≤ n − 1, a = a − a . What is the sum of the absolute n, 0 n,n n,k n − 1 ,k n − 1 ,k − 1 values of all numbers in the 2018th row?

解析

[ 10 ] Pascal has a triangle. In the n th row, there are n + 1 numbers a , a , a , . . . , a where n, 0 n, 1 n, 2 n,n a = a = 1. For all 1  k  n 1, a = a a . What is the sum of the absolute n, 0 n,n n,k n 1 ,k n 1 ,k 1 values of all numbers in the 2018th row? Proposed by: Michael Ren 2018 2 +2 Answer: 3 Let s be the sum of the absolute values of numbers in the n th row. For odd n , we have that n a , . . . , a alternate in sign as , + , , + , . . . , +, with the last term being a = 1. For even n, 1 n,n 1 n,n 1 n , we have that a , . . . , a alternate in sign as , + , , + , . . . , +, and a = 0. These facts can n, 1 n,n 2 n,n 1 n 1 be proven by induction. Thus, s = 1 a + a · · · + ( 1) a + 1. Applying the recursion, n n, 1 n, 2 n,n 1 n 1 for n > 0 this becomes s = 1 ( a a ) + ( a a ) · · · + ( 1) ( a n n 1 , 1 n 1 , 0 n 1 , 2 n 1 , 1 n 1 ,n 1 n 2 n 1 a ) + 1 = 2(1 a + a · · · + ( 1) a + 1) 1 + ( 1) . In other words, if n n 1 ,n 2 n 1 , 1 n 1 , 2 n 1 ,n 2 is even then s = 2 s 2 and if n is odd then s = 2 s . This means that s = 4 s 2. Since n n 1 n n 1 2 n 2 n 2 2018 2017 2015 2018 is even, we can write s = 4 s 2 = 2 2 2 · · · 2. Applying the formula 2018 2016 2019 2018 2018 2 2 2 +2 for the sum of a geometric series, we get s = 2 = . 2018 4 1 3