题目 已知 $S(n,k)=\displaystyle \sum_{i=1}^n{i^k}$，其中 $k,n\in\mathbb N^{\ast}$． 问题1 求 $S(n,1)$，$S(n,2)$，$S(n,3)$； 答案1 $S(n,1)=\dfrac 12n^2+\dfrac 12n$，$S(n,2)=\dfrac 13n^3+\dfrac 12n^2+\dfrac 16n$，$S(n,3)=\dfrac 14n^4+\dfrac 12n^3+\dfrac 14n^2$ 解析1 根据等差数列的求和公式，有\[S(n,1)=\dfrac 12n(n+1)=\dfrac 12n^2+\dfrac 12n.\]由和立方公式，有\[(i+1)^3=i^3+3i^2+3i+1,\]于是\[\sum_{i=1}^n(i+1)^3=\sum_{i=1}^ni^3+3S(n,2)+3S(n,1)+n,\]即\[(n+1)^3-n-1=3S(n,2)+3S(n,1),\]整理可得\[S(n,2)=\dfrac 13n^3+\dfrac 12n^2+\dfrac 16n.\]类似的，由\[(i+1)^4=i^4+4i^3+6i^2+4i+1,\]可得\[S(n,3)=\dfrac 14n^4+\dfrac 12n^3+\dfrac 14n^2.\] 备注1 问题2 给出 $S(n,k)$ 关于 $k$ 的一个递推公式． 答案2 $S(n,k)=\dfrac{(n+1)^{k+1}-n-1-{\rm C}_{k+1}^{k-1}S(n,k-1)-{\rm C}_{k+1}^{k-2}S(n,k-2)-\cdots-{\rm C}_{k+1}^1S(n,1)}{k+1}$ 解析2 由二项式定理，有\[(i+1)^{k+1}=i^{k+1}+{\rm C}_{k+1}^ki^k+{\rm C}_{k+1}^{k-1}i^{k-1}+\cdots+{\rm C}_{k+1}^1i+1,\]于是\[\sum_{i=1}^n(i+1)^{k+1}=\sum_{i=1}^ni^{k+1}+{\rm C}_{k+1}^kS(n,k)+{\rm C}_{k+1}^{k-1}S(n,k-1)+\cdots+{\rm C}_{k+1}^1S(n,1)+n,\]整理可得\[S(n,k)=\dfrac{(n+1)^{k+1}-n-1-{\rm C}_{k+1}^{k-1}S(n,k-1)-{\rm C}_{k+1}^{k-2}S(n,k-2)-\cdots-{\rm C}_{k+1}^1S(n,1)}{k+1}.\] 备注2 事实上，有\[S(n,k)=\dfrac{1}{k+1}\sum_{i=1}^{k+1}{\rm C}_{k+1}^iB_{k+1-i}n^i,\]其中\[\dfrac{z}{{\rm e}^z-1}=\sum_{n=0}^{+\infty}B_n\dfrac{z^n}{n!}.\]中间用到的 $B_n$ 被称为伯努利数（Bernoulli number）．<br/>观察下列等式：\begin{align*}<br/>\displaystyle\sum_{i=1}^{n}i&=\dfrac{1}{2}n^2+\dfrac{1}{2}n,\\<br/>\displaystyle\sum_{i=1}^{n}i^2&=\dfrac{1}{3}n^3+\dfrac{1}{2}n^2+\dfrac{1}{6}n,\\<br/>\displaystyle\sum_{i=1}^{n}i^3&=\dfrac{1}{4}n^4+\dfrac{1}{2}n^3+\dfrac{1}{4}n^2,\\<br/>\displaystyle\sum_{i=1}^{n}i^4&=\dfrac{1}{5}n^5+\dfrac{1}{2}n^4+\dfrac{1}{3}n^3-\dfrac{1}{30}n,\\<br/>\displaystyle\sum_{i=1}^{n}i^5&=\dfrac{1}{6}n^6+\dfrac{1}{2}n^5+\dfrac{5}{12}n^4-\dfrac{1}{12}n^2,\\<br/>\displaystyle\sum_{i=1}^{n}i^6&=\dfrac{1}{7}n^7+\dfrac{1}{2}n^6+\dfrac{1}{2}n^5-\dfrac{1}{6}n^3+\dfrac{1}{42}n,\\<br/>&\vdots\\<br/>\displaystyle\sum_{i=1}^{n}i^k&=a_{k+1}n^{k+1}+a_kn^k+a_{k-1}n^{k-1}+a_{k-2}n^{k-2}+\cdots+a_1n+a_0,<br/>\end{align*}可以推测，当 $k \geqslant 2 \left(k\in\mathbb{N}^{*}\right)$ 时，$$a_{k+1}=\dfrac{1}{k+1} , a_k=\dfrac{1}{2},$$试写出 $a_{k-1}$，$a_{k-2}$．<br/>这道题考查合情推理，观察出答案很容易：$a_{k-1}=\dfrac{k}{12}$，$a_{k-2}=0$．<br/>下面用数学归纳法给出证明．<br/>为表述方便，设\[<br/>S_k(n)=\displaystyle\sum_{i=1}^{n}i^k=a_{k+1}^{(k)}n^{k+1}+a_k^{(k)}n^k+a_{k-1}^{(k)}n^{k-1}+a_{k-2}^{(k)}n^{k-2}+\cdots+a_1^{(k)}n+a_0^{(k)},<br/>\]<case>归纳基础</case>我们现在要证明对任意 $k \geqslant 2 \left(k\in\mathbb{N}^{*}\right)$，都有\[\begin{cases}<br/>a_{k+1}^{(k)}=\dfrac{1}{k+1},\\<br/>a_k^{(k)}=\dfrac{1}{2},\\<br/>a_{k-1}^{(k)}=\dfrac{k}{12},\\<br/>a_{k-2}^{(k)}=0.<br/>\end{cases}\]由题意，当 $k=2,3,4$ 时，欲证等式均成立．<br/><case>递推证明</case>假设欲证等式对 $k=m-1,m,m+1$ 均成立，则有\begin{align*}<br/>S_{m+1}(n)&=\dfrac{1}{m+2}n^{m+2}+\dfrac{1}{2}n^{m+1}+\dfrac{m+1}{12}n^{m}+\cdots+a_0^{(m+1)},\\<br/>S_{m}(n)&=\dfrac{1}{m+1}n^{m+1}+\dfrac{1}{2}n^{m}+\cdots+a_0^{(m)},\\<br/>S_{m-1}(n)&=\dfrac{1}{m}n^m+\cdots+a_0^{(m-1)},<br/>\end{align*}因此 $k=m+2$ 时，由递推公式可知\begin{align*}<br/>a_{m+3}^{(m+2)}&=\dfrac{1}{m+3},\\<br/>a_{m+2}^{(m+2)}&=\dfrac{1}{m+3}\left(\mathrm{C}_{m+3}^{1}-\dfrac{\mathrm{C}_{m+3}^{2}}{m+2}\right)=\dfrac{1}{2},\\<br/>a_{m+1}^{(m+2)}&=\dfrac{1}{m+3}\left(\mathrm{C}_{m+3}^{2}-\dfrac{\mathrm{C}_{m+3}^{2}}{2}-\dfrac{\mathrm{C}_{m+3}^{3}}{m+1}\right)=\dfrac{m+2}{12},\\<br/>a_{m+1}^{(m+2)}&=\dfrac{1}{m+3}\left(\mathrm{C}_{m+3}^{3}-\dfrac{(m+1)\mathrm{C}_{m+3}^{2}}{12}-\dfrac{\mathrm{C}_{m+3}^{3}}{2}-\dfrac{\mathrm{C}_{m+3}^{4}}{m}\right)=0,<br/>\end{align*}故 $k=m+2$ 时，欲证等式也成立．<br/>综上所述，欲证等式对任意 $k \geqslant 2 \left(k\in\mathbb{N}^{*}\right)$ 都成立．