HMMT 十一月 2018 · 冲刺赛 · 第 35 题

HMMT November 2018 — Guts Round — Problem 35

[ 20 ] 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 . Let N be the value of the sum n, 0 n,n n,k n − 1 ,k n − 1 ,k − 1 2018 ∑ | a | 2018 ,k ( ) . 2018 k k =0 Estimate N . −| N − E | / 70 An estimate of E > 0 earns b 20 · 2 c points.

解析

[ 20 ] 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 . Let N be the value of the sum n, 0 n,n n,k n 1 ,k n 1 ,k 1 2018 X | a | 2018 ,k . 2018 k k =0 Estimate N . | N E | / 70 An estimate of E > 0 earns b 20 · 2 c points. Proposed by: Michael Ren Answer: 780 . 9280674537 A good estimate for this question is to use the fact that 2018 2018 X 2 + 2 | a | = , 2018 ,k 3 k =0 ✓ ◆ 2018 1 the answer to Guts 17. This suggests that each | a | is roughly of its corresponding entry 2018 ,k 3 k 2018 in the usual Pascal’s triangle, as the sum of the terms in the 2018th row of Pascal’s triangle is 2 . 2018 This then gives an estimate of , which earns 6 points. Code for computing answer in Python 3: 3 import math lists=[[1]] for i in range(2018): newlist=[] for j in range(i): newlist.append(lists[-1][j+1]-lists[-1][j]) lists.append([1]+newlist+[1]) big=math.factorial(2018) sum=0 for i in range(2019): sum+=abs(lists[-1][i])/(big//math.factorial(i)//math.factorial(2018-i)) print(sum)