题目

Problem

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

$\sum_{r=1}^n r(r+1)(r+5) = \frac{1}{4}n(n+a)(n+b)(n+c)$

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

(5)

(b) Hence determine the value of

$20 \times 21 \times 25 + 21 \times 22 \times 26 + \dots + 40 \times 41 \times 45$

(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 $\sum_{r=1}^n r(r+1)(r+5) = \frac{1}{4}n(n+a)(n+b)(n+c)$

其中 $a, b, c$ 为待求的整数。

(b) 由此确定以下算式的值： $20 \times 21 \times 25 + 21 \times 22 \times 26 + \dots + 40 \times 41 \times 45$

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,

r(r+1)(r+5)=r^3+6r^2+5r.

Using the standard summation formulae,

\begin{align*} \sum_{r=1}^{n}r(r+1)(r+5) =&\,\sum_{r=1}^{n}r^3 +6\sum_{r=1}^{n}r^2 +5\sum_{r=1}^{n}r\\ =&\,\frac{n^2(n+1)^2}{4} +n(n+1)(2n+1)\\ &\,\hspace{2pt}+\frac52n(n+1)\\ =&\,\frac14n(n+1) \bigl[n(n+1)+4(2n+1)+10\bigr]\\ =&\,\frac14n(n+1)(n^2+9n+14)\\ =&\,\frac14n(n+1)(n+2)(n+7). \end{align*}

Thus the required result is

\boxed{ \sum_{r=1}^{n}r(r+1)(r+5) =\frac14n(n+1)(n+2)(n+7)},

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 +21\cdot22\cdot26+\cdots+40\cdot41\cdot45\\ =&\,\sum_{r=20}^{40}r(r+1)(r+5)\\ =&\,\sum_{r=1}^{40}r(r+1)(r+5) -\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*}