Phantom Bird

外观

FFMP 20250101

约 863 字大约 3 分钟

2024-12-28

新年快乐！

20252025 是一个神奇的年份，比如它是 45245^2，也是 ∑i=19i3\sum\limits_{i=1}^{9}i^3.

我们设定义域为 N\mathbb{N} 的函数 Fn(a)=∑i=1ain (n∈N)F_n(a)=\sum\limits_{i=1}^a i^n \ (n \in \mathbb{N})，则 20252025 可以表示为 F3(9)F_3(9).

  1. 求证：∀a∈N,F3(a)\forall a\in\mathbb{N}, F_3(a) 是完全平方数；

  2. 若 F3(a)F2(a)∈Z{\dfrac{F_3(a)}{F_2(a)}} \in \mathbb{Z}，求 aa 的值；

  3. 求证：若 n>1,Fn(a)n>1, F_n(a) 的 nn 次项系数为 12\dfrac{1}{2} ；

  4. 求 Fn(a)F_n(a) 关于 F0(a),F1(a),⋯ ,Fn−1(a)F_0(a), F_1(a), \cdots, F_{n-1}(a) 的表达式.

提示

求 F2(a)F_2(a) 的解析式：

(a+1)3−a3=3a2+3a+1(a+1)3−13=∑i=1a(a+1)3−a3=∑i=1a3a2+3a+1=∑i=1a3a2+∑i=1a3a+∑i=1a1=3F2(a)+3a(a+1)2+a整理可得 F2(a)=a(a+1)(2a+1)6\begin{align*} (a+1)^3-a^3&=3a^2+3a+1 \\ (a+1)^3-1^3&=\sum_{i=1}^{a} (a+1)^3-a^3 \\ &= \sum_{i=1}^{a} 3a^2+3a+1 \\ &= \sum_{i=1}^{a} 3a^2+ \sum_{i=1}^{a}3a+ \sum_{i=1}^{a}1 \\ &= 3F_2(a)+\frac{3a(a+1)}{2}+a \\ \text{整理可得 } F_2(a) &= \frac{a(a+1)(2a+1)}{6} \end{align*}

答案


  1. 写出 F3(a)F_3(a) 的前几项，猜想 F3(a)=[a(a+1)2]2F_3(a) = \left[\dfrac{a(a+1)}{2}\right]^2.
    证明：使用数学归纳法，
    ① F3(1)=[1×(1+1)2]2=1F_3(1) = \left[\dfrac{1\times(1+1)}{2}\right]^2 = 1；
    ② 若 F3(a)=[a(a+1)2]2F_3(a) = \left[\dfrac{a(a+1)}{2}\right]^2， F3(a+1)=a2(a+1)24+(a+1)3=(a+2)2(a+1)24F_3(a+1) = {\dfrac{a^2(a+1)^2}{4}} + (a+1)^3 = {\dfrac{(a+2)^2(a+1)^2}{4}}.
    显然 a(a+1)2\dfrac{a(a+1)}{2} 为整数，证毕。

设 F3(a)F2(a)=a2(a+1)2/4a(a+1)(2a+1)/6=3a(a+1)2(2a+1)=k则 3a(a+1)=2k(2a+1)整理得 3a2+(3−4k)a−2k=0解得 a=4k−3±16k2+96设 16k2+9=t∈N则有 (t+4k)(t−4k)=9由 9=9×1=3×3(t,k≥0)得 k=0 或 1所以 a=1(0,−1,23 舍去)\begin{align*} \text{设 }& \frac{F_3(a)}{F_2(a)} = \frac{a^2(a+1)^2/4}{a(a+1)(2a+1)/6} = \frac{3a(a+1)}{2(2a+1)}=k \\ \text{则 }& 3a(a+1) = 2k(2a+1) \\ \text{整理得 }& 3a^2 + (3-4k)a - 2k = 0 \\ \text{解得 }& a = \frac{4k-3 \pm \sqrt{16k^2+9}}{6} \\ \text{设 }& \sqrt{16k^2+9}=t \in \mathbb{N} \\ \text{则有 }& (t+4k)(t-4k)=9 \\ \text{由 }& 9 = 9\times1 = 3\times3 \quad(t,k\ge0) \\ \text{得 }& k=0 \text{ 或 } 1 \\ \text{所以 }& a=1 \quad(0,-1,\frac{2}{3} \text{ 舍去}) \end{align*}


  1. 显然计算过程中可以省略低次项。于是

(a+1)n+1=an+1+(n+1)an+n(n+1)2an−1+O(an−2)(a+1)n+1−an+1=(n+1)an+n(n+1)2an−1+O(an−2)(a+1)n+1−1n+1=∑i=1a(a+1)n+1−an+1=∑i=1a(n+1)in+n(n+1)2an−1+O(in−2)=(n+1)Fn(a)+n(n+1)2Fn−1(a)+O(an−2)整理可得 Fn(a)=(a+1)n+1n+1−n2Fn−1(a)+O(an−1)=an+1n+1+an−n2Fn−1(a)+O(an−1)不难发现 Fn(a) 的 n 次项为 1n,故 n−1 次项为 1−n2×1n=12，证毕.\begin{gather*} \begin{aligned} (a+1)^{n+1} &= a^{n+1} + (n+1)a^n + \frac{n(n+1)}{2}a^{n-1} + O(a^{n-2}) \\ (a+1)^{n+1}-a^{n+1} &= (n+1)a^n+\frac{n(n+1)}{2}a^{n-1} + O(a^{n-2}) \\ (a+1)^{n+1}-1^{n+1} &= \sum_{i=1}^{a} (a+1)^{n+1}-a^{n+1} \\ &= \sum_{i=1}^{a} (n+1) i^n+\frac{n(n+1)}{2}a^{n-1} + O(i^{n-2}) \\ &= (n+1)F_n(a)+\frac{n(n+1)}{2}F_{n-1}(a)+O(a^{n-2}) \\ \text{整理可得 } F_n(a) &= \frac{(a+1)^{n+1}}{n+1}-\frac{n}{2}F_{n-1}(a) + O(a^{n-1}) \\ &= \frac{a^{n+1}}{n+1} + a^n-\frac{n}{2}F_{n-1}(a) + O(a^{n-1}) \end{aligned} \\ \text{不难发现 } F_n(a) \text{ 的 } n \text{ 次项为 } \frac{1}{n}, \\ \text{故 } n-1 \text{ 次项为 } 1-\frac{n}{2}\times\frac{1}{n} = \frac{1}{2}\text{，证毕.} \end{gather*}

(a+1)n+1=∑i=0n+1Cn+1iai(a+1)n+1−an+1=∑i=0nCn+1iai(a+1)n+1−1n+1=∑w=1a∑i=0nCn+1iwi(a+1)n+1−1=∑w=1a∑i=0nCn+1iwi=∑w=1a(Cn+1nwn+∑i=0n−1Cn+1iwi)=(∑w=1aCn+1nwn)+(∑w=1a∑i=0n−1Cn+1iwi)=(n+1)(∑w=1awn)+(∑i=0n−1∑w=1aCn+1iwi)=(n+1)(∑w=1awn)+(∑i=0n−1Cn+1i∑w=1awi)=(n+1)Fn(a)+∑i=0n−1Cn+1iFi(a)Fn(a)=1n+1[(a+1)n+1−1−∑i=0n−1Cn+1iFi(a)]\begin{gather*} \begin{aligned} (a+1)^{n+1} &= \sum_{i=0}^{n+1} C_{n+1}^i a^i \\ (a+1)^{n+1}-a^{n+1} &= \sum_{i=0}^{n} C_{n+1}^i a^i \\ (a+1)^{n+1}-1^{n+1} &= \sum_{w=1}^{a} \sum_{i=0}^{n} C_{n+1}^i w^i \\ \end{aligned} \\ \begin{aligned} (a+1)^{n+1}-1 &= \sum_{w=1}^{a} \sum_{i=0}^{n} C_{n+1}^i w^i \\ &= \sum_{w=1}^{a} \left(C_{n+1}^{n}w^{n}+\sum_{i=0}^{n-1} C_{n+1}^i w^i \right) \\ &= \left( \sum_{w=1}^{a} C_{n+1}^{n}w^{n} \right) + \left( \sum_{w=1}^{a}\sum_{i=0}^{n-1} C_{n+1}^i w^i \right) \\ &= (n+1)\left( \sum_{w=1}^{a} w^{n} \right) + \left( \sum_{i=0}^{n-1}\sum_{w=1}^{a} C_{n+1}^i w^i \right) \\ &= (n+1)\left( \sum_{w=1}^{a} w^{n} \right) + \left( \sum_{i=0}^{n-1} C_{n+1}^i \sum_{w=1}^{a} w^i \right) \\ &= (n+1)F_n(a) + \sum_{i=0}^{n-1} C_{n+1}^i F_i(a) \\ \end{aligned} \\ F_n(a) = \frac{1}{n+1} \left[ (a+1)^{n+1} - 1 - \sum_{i=0}^{n-1} C_{n+1}^i F_i(a) \right] \end{gather*}

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