HMMT 二月 2011 · 冲刺赛 · 第 12 题
HMMT February 2011 — Guts Round — Problem 12
题目详情
- [ 8 ] A sequence of integers { a } is defined as follows: a = i for all 1 ≤ i ≤ 5, and a = a a · · · a − 1
i i i 1 2 i − 1
2011
∑
for all i > 5. Evaluate a a · · · a − a .2
1 2 2011
i =1 i
14 HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUNDthb + c − a
英文原题
[ 8 ] A sequence of integers { a i } is defined as follows: a i = i for all 1 ≤ i ≤ 5, and a i = a 1 a 2 · · · a i − 1 − 1
for all i > 5. Evaluate a 1 a 2 · · · a 2011 −
2011 ∑
i =1
a 2
i .
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14 th HARVARD-MIT MATHEMATICS TOURNAMENT, 12 FEBRUARY 2011 — GUTS ROUND
解析
英文解析
- [ 8 ] A sequence of integers { a } is defined as follows: a = i for all 1 ≤ i ≤ 5, and a = a a · · · a − 1
i i i 1 2 i − 1
2011
∑
for all i > 5. Evaluate a a · · · a − a .2
1 2 2011
i =1 i
Answer: − 1941
For all i ≥ 6, we have a = a a · · · a − 1. Soi 1 2 i − 1
a = a a · · · a − 1
i +1 1 2 i = ( a a · · · a ) a − 1
1 2 i − 1 i = ( a + 1) a − 1
i i = a + a − 1 .2
2 ii
Therefore, for all i ≥ 6, we have a = a − a + 1, and we obtain thati +1 i
2011 i
∑
a a · · · a − a 2
1 2 2011
i =1 i
5 2011
∑ ∑
2 2 = a + 1 − a − a
2012
i ii =1 i =6
5 2011
∑ ∑
= a + 1 − i − ( a − a + 1)2
2012 i +1 ii =1 i =6 = a + 1 − 55 − ( a − a + 2006)
2012 2012 6 = a − 2060 = − 19416
b + c − a