Answer

问题及解答

证明: $\sum\limits_{i=1}^{n}i^3=(\frac{n(n+1)}{2})^2$.

Posted by haifeng on 2019-12-20 20:58:32 last update 2019-12-21 20:40:33 | Edit | Answers (2)

证明: $\sum\limits_{i=1}^{n}i^3=(\frac{n(n+1)}{2})^2$.

 

至少有三种证法.

(1) 数学归纳法(mathematical induction)

(2) 考虑telescoping sum $\sum\limits_{i=1}^{n}[(i+1)^4-i^4]$

(3) 由 Abu Bekr Mohammed ibn Alhusain Alkarchi 在公元后大约1010年左右证明. 使用一个边长为 $\frac{n(n+1)}{2}$ 的正方形来证明. 此正方形的边被分割成长度依次为 $1,2,3,\ldots,n$ 的线段.

 


(4) 第四种证明是由 David Chen (陈)的儿子(六年级)在2019年12月20日发现的. 使用了下面的观察, 也是一种 telescoping sum

首先 $i^3=(i-1)i(i+1)+i$, 然后观察到

\[
(i-1)i(i+1)=\frac{1}{4}\Bigl[(i-1)i(i+1)(i+2)-(i-2)(i-1)i(i+1)\Bigr]
\]

Note: $(i-1)i(i+1)=i(i^2-1)$ 这个形式也出现在 $y^2=x^3+x^2$ 的有理参数化中. (令 $x=t^2-1$, 则 $y=t(t^2-1)$)

代数几何中, 平面三次曲线族 $y^2=x(x-1)(x-\lambda)$ (这里 $\lambda\in\mathbb{R}$) 只有对于 $\lambda=0$ 或 $\lambda=1$ 时能被有理参数化. 

参见 Klaus Hulek 著, 胥鸣伟 译 《初等代数几何》P. 6-7, 例0.5.

 

References of (3)

James Stewart, Calculus (7th Edition) Appendix E. Exercise 40.

 

顺便考虑一下 $\sum\limits_{i=1}^{n}i^4$,  $\sum\limits_{i=1}^{n}i^5$ 等等.

1

Posted by haifeng on 2019-12-20 22:51:07

使用第二种方法, 即考虑 telescoping sum

\[
\sum_{i=1}^{n}\bigl[(i+1)^5-i^5\bigr]
\]

可以证明

\[
\sum_{i=1}^{n}i^4=\frac{n}{30}(6n^4+15n^3+10n^2-1)
\]

 

Pf.  

\[
\sum_{i=1}^{n}\bigl[(i+1)^5-i^5\bigr]=\sum_{i=1}^{n}(i+1)^5-\sum_{i=1}^{n}i^5=\sum_{j=2}^{n+1}j^5-\sum_{j=1}^{n}j^5=(n+1)^5-1
\]

另一方面, 记 $S=\sum\limits_{i=1}^{n}i^4$,

\[
\begin{split}
\sum_{i=1}^{n}\bigl[(i+1)^5-i^5\bigr]&=\sum_{i=1}^{n}\bigl[i^5+5i^4+10i^3+10i^2+5i+1-i^5\bigr]\\
&=5\sum_{i=1}^{n}i^4+10\sum_{i=1}^{n}i^3+10\sum_{i=1}^{n}i^2+5\sum_{i=1}^{n}i+\sum_{i=1}^{n}1\\
&=5S+10\cdot(\frac{n(n+1)}{2})^2+10\cdot\frac{n(n+1)(2n+1)}{6}+5\cdot\frac{n(n+1)}{2}+n\\
&=5S+\frac{5}{2}n^2(n+1)^2+\frac{5}{3}(n^2+n)(2n+1)+\frac{5}{2}(n^2+n)+n\\
&=5S+\frac{5}{2}n^2(n^2+2n+1)+\frac{5}{3}(2n^3+3n^2+n)+\frac{5}{2}n^2+\frac{7}{2}n\\
&=5S+\frac{5}{2}(n^4+2n^3+n^2)+\frac{10}{3}n^3+5n^2+\frac{5}{3}n+\frac{5}{2}n^2+\frac{7}{2}n\\
&=5S+\frac{5}{2}n^4+\frac{25}{3}n^3+10n^2+\frac{31}{6}n
\end{split}
\]

因此

\[
\begin{split}
&(n+1)^5-1=5S+\frac{5}{2}n^4+\frac{25}{3}n^3+10n^2+\frac{31}{6}n\\
\Rightarrow&n^5+5n^4+10n^3+10n^2+5n=5S+\frac{5}{2}n^4+\frac{25}{3}n^3+10n^2+\frac{31}{6}n\\
\Rightarrow&n^5+\frac{5}{2}n^4+\frac{5}{3}n^3-\frac{1}{6}n=5S\\
\Rightarrow&S=\frac{1}{30}n(6n^4+15n^3+10n^2-1)
\end{split}
\]

 

2

Posted by haifeng on 2019-12-21 10:00:45

下面给出第四种证法.

首先 $i^3=(i-1)i(i+1)+i$, 然后观察到

\[
(i-1)i(i+1)=\frac{1}{4}\Bigl[(i-1)i(i+1)(i+2)-(i-2)(i-1)i(i+1)\Bigr]
\]

于是

\[
\begin{split}
\sum_{i=1}^{n}i^3&=\sum_{i=1}^{n}\bigl[(i-1)i(i+1)+i\bigr]\\
&=\sum_{i=1}^{n}(i-1)i(i+1)+\sum_{i=1}^{n}i\\
&=\sum_{i=1}^{n}\frac{1}{4}\Bigl[(i-1)i(i+1)(i+2)-(i-2)(i-1)i(i+1)\Bigr]+\frac{n(n+1)}{2}\\
&=\frac{1}{4}\Bigl[\sum_{i=1}^{n}(i-1)i(i+1)(i+2)-\sum_{i=1}^{n}(i-2)(i-1)i(i+1)\Bigr]+\frac{n(n+1)}{2}\\
&=\frac{1}{4}\Bigl[\sum_{i=1}^{n}(i-1)i(i+1)(i+2)-\sum_{j=0}^{n-1}(j-1)j(j+1)(j+2)\Bigr]+\frac{n(n+1)}{2}\\
&=\frac{1}{4}\Bigl[\sum_{i=1}^{n}(i-1)i(i+1)(i+2)-\sum_{i=0}^{n-1}(i-1)i(i+1)(i+2)\Bigr]+\frac{n(n+1)}{2}\\
&=\frac{1}{4}\Bigl[(n-1)n(n+1)(n+2)-(-1)\cdot 0\cdot 1\cdot 2\Bigr]+\frac{n(n+1)}{2}\\
&=\frac{1}{4}(n-1)n(n+1)(n+2)+\frac{n(n+1)}{2}\\
&=\frac{1}{4}n(n+1)\bigl[(n-1)(n+2)+2\bigr]\\
&=(\frac{n(n+1)}{2})^2
\end{split}
\]