HMMT 二月 2013 · 代数 · 第 7 题
HMMT February 2013 — Algebra — Problem 7
题目详情
英文原题
Compute ∞/summationdisplay a1=0∞/summationdisplay a2=0· · ·∞/summationdisplay a7=0a1+a2+· · ·+a7 3a1+a2+···+a7.
解析
英文解析
- Compute
∞ ∞ ∞
∑ ∑ ∑
a + a + · · · + a
1 2 7
· · · .
a + a + ··· + a
1 2 7
a =0 a =0 a =03
1 2 7
Answer: 15309/256 Note that, since this is symmetric in a through a ,
1 7
∞ ∞ ∞ ∞ ∞ ∞
∑ ∑ ∑ ∑ ∑ ∑
a + a + · · · + a a
1 2 7 1
· · · = 7 · · ·
a + a + ··· + a a + a + ··· + a
1 2 7 1 2 7
3 3
a =0 a =0 a =0 a =0 a =0 a =0
1 2 7 1 2 7
( ) ( )
∞ ∞6
∑ ∑
a 1 = 7 .1
a a
3 31
a =0 a =0
( )1
∑ ∑
a 1 3 3 153096
If S = , then 3 S − S = = 3 / 2, so S = 3 / 4. It follows that the answer equals 7 · · = .
a a
3 3 4 2 256
∑ ∑ ∑
∞ ∞ ∞
a + a + ··· + a
1 2 7
Alternatively, let f ( z ) = · · · z . Note that we can rewrite f ( z ) =
a =0 a =0 a =0
1 2 7
∑ ∑ ∑ ∑
∞ ∞ ∞ ∞
a 7 1 ′
( z ) = . Furthermore, note that zf ( z ) = · · · ( a + a + · · · +
7 1 2
a =0 a =0 a =0 a =0
(1 − z )
1 2 7
′
f (1 / 3)
a + a + ··· + a ′7
1 2 7
a ) z , so the sum in question is simply . Since f ( x ) = , it follows that the sum
7 8
3 (1 − z )
7 · 3 153097
is equal to = .
2 2568
2 2 4 4 6 6
x + y x + y x + y