题目

Problem

(a) Use the standard results for summations to show that for all positive integers $n$

\begin{align*} &\,\sum_{r=1}^n r(r+1)(r+5)\\ =&\,\frac14n(n+a)\\ &\,\hspace{2pt}\times(n+b)(n+c) \end{align*}

where $a, b$ and $c$ are integers to be determined.

(5)

(b) Hence determine the value of

\begin{align*} &\,20\times21\times25\\ &\,\hspace{2pt}+21\times22\times26+\cdots\\ &\,\hspace{4pt}+40\times41\times45 \end{align*}

(2)

题目中文翻译

(a) 利用标准的求和公式，证明对所有正整数 $n$ ，均有

\begin{align*} &\,\sum_{r=1}^n r(r+1)(r+5)\\ =&\,\frac14n(n+a)\\ &\,\hspace{2pt}\times(n+b)(n+c) \end{align*}

其中 $a, b, c$ 为待求的整数。

(5)

(b) 由此确定以下算式的值：

\begin{align*} &\,20\times21\times25\\ &\,\hspace{2pt}+21\times22\times26+\cdots\\ &\,\hspace{4pt}+40\times41\times45 \end{align*}

(2)

解答

展开

先展开一般项，再分别使用立方和、平方和与自然数和公式。提取公因子 $\frac14n(n+1)$ 后，将剩余二次式因式分解，便可与题目给出的形式比较。

展开

First,

\begin{align*} &\,r(r+1)(r+5)\\ =&\,r^3+6r^2+5r. \end{align*}

Using the standard summation formulae,

\begin{align*} &\,\sum_{r=1}^{n}r(r+1)(r+5)\\ =&\,\sum_{r=1}^{n}r^3\\ &\,\hspace{2pt}+6\sum_{r=1}^{n}r^2\\ &\,\hspace{4pt}+5\sum_{r=1}^{n}r\\ =&\,\frac{n^2(n+1)^2}{4}\\ &\,\hspace{2pt}+n(n+1)(2n+1)\\ &\,\hspace{2pt}+\frac52n(n+1)\\ =&\,\frac14n(n+1)\\ &\,\hspace{2pt}\times\big[n(n+1)\\ &\,\hspace{4pt}+4(2n+1)+10\big]. \end{align*}

The expression in brackets simplifies to

\begin{align*} &\,n(n+1)\\ &\,\hspace{2pt}+4(2n+1)+10\\ =&\,n^2+n+8n+4+10\\ =&\,n^2+9n+14\\ =&\,(n+2)(n+7). \end{align*}

Therefore,

\begin{align*} &\,\sum_{r=1}^{n}r(r+1)(r+5)\\ =&\,\frac14n(n+1)\\ &\,\hspace{2pt}\times(n^2+9n+14)\\ =&\,\frac14n(n+1)\\ &\,\hspace{2pt}\times(n+2)(n+7), \end{align*}

with, for example,

\boxed{a=1,\quad b=2,\quad c=7}.

展开

给定算式的一般项是 $r(r+1)(r+5)$ ，范围从 $r=20$ 到 $r=40$ 。承接 (a)，用前 40 项的和减去前 19 项的和。

展开

Using the result from part (a),

\begin{align*} &\,20\cdot21\cdot25 \\ &\,\hspace{2pt}+21\cdot22\cdot26+\cdots\\ &\,\hspace{4pt}+40\cdot41\cdot45\\ =&\,\sum_{r=20}^{40}r(r+1)(r+5)\\ =&\,\sum_{r=1}^{40}r(r+1)(r+5)\\ &\,\hspace{2pt}-\sum_{r=1}^{19}r(r+1)(r+5)\\ =&\,\frac14(40)(41)(42)(47)\\ &\,\hspace{2pt}-\frac14(19)(20)(21)(26)\\ =&\,809340-51870\\ =&\,\boxed{757470}. \end{align*}