新年快乐！

$2025$ 是一个神奇的年份，比如它是 $45^2$，也是 $\sum\limits_{i=1}^{9}i^3$.

我们设定义域为 $\mathbb{N}$ 的函数 $F_n(a)=\sum\limits_{i=1}^a i^n \ (n \in \mathbb{N})$，则 $2025$ 可以表示为 $F_3(9)$.

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

2. 若 ${\dfrac{F_3(a)}{F_2(a)}} \in \mathbb{Z}$，求 $a$ 的值；

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

4. 求 $F_n(a)$ 关于 $F_0(a), F_1(a), \cdots, F_{n-1}(a)$ 的表达式.

提示

求 $F_2(a)$ 的解析式：

$$\begin{align*} (a+1)^3-a^3&=3a^2+3a+1 \\ (a+1)^3-1^3&=\sum_{i=1}^{a} (a+1)^3-a^3 \\ &= \sum_{i=1}^{a} 3a^2+3a+1 \\ &= \sum_{i=1}^{a} 3a^2+ \sum_{i=1}^{a}3a+ \sum_{i=1}^{a}1 \\ &= 3F_2(a)+\frac{3a(a+1)}{2}+a \\ \text{整理可得 } F_2(a) &= \frac{a(a+1)(2a+1)}{6} \end{align*}$$

答案

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

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

3. 
   显然计算过程中可以省略低次项。于是

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

$$\begin{gather*} \begin{aligned} (a+1)^{n+1} &= \sum_{i=0}^{n+1} C_{n+1}^i a^i \\ (a+1)^{n+1}-a^{n+1} &= \sum_{i=0}^{n} C_{n+1}^i a^i \\ (a+1)^{n+1}-1^{n+1} &= \sum_{w=1}^{a} \sum_{i=0}^{n} C_{n+1}^i w^i \\ \end{aligned} \\ \begin{aligned} (a+1)^{n+1}-1 &= \sum_{w=1}^{a} \sum_{i=0}^{n} C_{n+1}^i w^i \\ &= \sum_{w=1}^{a} \left(C_{n+1}^{n}w^{n}+\sum_{i=0}^{n-1} C_{n+1}^i w^i \right) \\ &= \left( \sum_{w=1}^{a} C_{n+1}^{n}w^{n} \right) + \left( \sum_{w=1}^{a}\sum_{i=0}^{n-1} C_{n+1}^i w^i \right) \\ &= (n+1)\left( \sum_{w=1}^{a} w^{n} \right) + \left( \sum_{i=0}^{n-1}\sum_{w=1}^{a} C_{n+1}^i w^i \right) \\ &= (n+1)\left( \sum_{w=1}^{a} w^{n} \right) + \left( \sum_{i=0}^{n-1} C_{n+1}^i \sum_{w=1}^{a} w^i \right) \\ &= (n+1)F_n(a) + \sum_{i=0}^{n-1} C_{n+1}^i F_i(a) \\ \end{aligned} \\ F_n(a) = \frac{1}{n+1} \left[ (a+1)^{n+1} - 1 - \sum_{i=0}^{n-1} C_{n+1}^i F_i(a) \right] \end{gather*}$$