AoPS 课后习题

题目来源：AoPS Vol 2

Chapter 15 Combinatorics

No. 248

Problem

For fixed \(n\), maximize the quantity \(\binom {2n+k} n \binom {2n-k} n\)。

Solution

TODO: 证明待补

\(k=0\) 时，原式取到最小值 \(\boxed{\binom {2n} {n} ^2}\)。

Chapter 16 Sequences and Series

No. 263

Problem

Evaluate \(\sum\limits_{i=1}^\infty \dfrac {i^2} {2^i}\).

Solution

把 \(\dfrac {k^2} {2^k}\) 拆成 \(k^2\) 个 \(\dfrac 1 {2^k}\) 相加。那么我们可以拉出 \(1\) 条从 \(\dfrac 1 2\) 开始的几何级数， \(3\) 条从 \(\dfrac 1 4\) 开始的几何级数，\(5\) 条从 \(\dfrac 1 8\) 开始的几何级数……

\[\begin{aligned} f(k) &= \sum\limits_{i=1}^\infty \dfrac {i^k} {2^i} \\ &= 2 \sum\limits_{i=1}^\infty \dfrac {i^k - (i-1)^k} {2^i} \end{aligned}\]

\[f(0) = 1 \]

\[f(1) = 2 \sum \dfrac 1 {2^i} = 2 f(0) = 2 \]

\[f(2) = 2 \sum \dfrac {2i-1} {2^i} = -2 + 4 \sum \dfrac {i} {2^i} = -2 + 4 f(1) = \boxed{6} \]

OI 闲话

\(f\) 有递推式：（边界 \(f_0 = 1\)）

\[f_n = 2 \sum\limits_{i=0}^{n-1} (-1)^{n-i+1} \binom n i f_i \]

搜索 https://oeis.org/A000629 得：

\[f_n = \sum\limits_{i=0}^{n} (-1)^{n-i} {n \brace i} \times i! \times 2^i \]

用 Poly 科技预处理第二类斯特林数·行，可做到 \(\Theta(n \log n)\)。

No. 277

Problem

Find \(\sum\limits_{x=0}^\infty \dfrac {\sin(x\alpha)} {3^x}\) if \(\sin \alpha = \dfrac 1 3\) and \(0 \le \alpha \le \dfrac \pi 2\).

Solution

令 \(z = \dfrac 2 9 \sqrt 2 + \dfrac 1 9 i\)。即模长为 \(\dfrac 1 3\) 辐角为 \(\alpha\) 的复数。

\[\begin{aligned} & \sum\limits_{x=0}^\infty \dfrac {\sin(x\alpha)} {3^x} \\ =& \sum\limits_{k=1}^\infty \text{Im } z^k \\ =& \text{Im } \sum\limits_{k=1}^\infty z^k \\ =& \text{Im} \left(\dfrac 1 {1 - z} - 1\right) \\ =& \boxed{\dfrac {5 + 2 \sqrt 2} {34}} \\ \end{aligned}\]

No. 278

Problem

数列 \(a\) 满足 \(a_i = a_1 + i - 1\)。已知若 \(a_i\) 为质数，则 \(i\) 为质数。求 \(a_{1993}\)。（题意有转化）

Solution

一眼看出 \(a_1 = 1\)，即 \(a_i = i\)。尝试用反证法构造证明没有其它解。

//TODO:

\(a_{1993} = \boxed{1993}\)。

Chapter 9 Complex Numbers

No. 151

Problem

Solution

No. 154

Problem

Solution