题目

Problem

Use the standard results for $\displaystyle\sum_{r=1}^{n} r^2$ and $\displaystyle\sum_{r=1}^{n} r^3$ to show that, for all positive integers $n$

$\sum_{r=1}^{n} r^2(r+2) = \frac{1}{12}n(n+1)(an^2 + bn + c)$

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

(4)

题目中文翻译

利用 $\displaystyle\sum_{r=1}^{n} r^2$ 和 $\displaystyle\sum_{r=1}^{n} r^3$ 的标准结果证明，对于所有正整数 $n$

$\sum_{r=1}^{n} r^2(r+2) = \frac{1}{12}n(n+1)(an^2 + bn + c)$

其中 $a$ 、 $b$ 和 $c$ 是待定整数。

解答

展开

先展开被加数 $r^2(r+2)=r^3+2r^2$ ，把原求和拆成三次方和与平方和。代入两个标准结果后提取公共因子 $\frac1{12}n(n+1)$ ，再整理括号内的二次式即可确定 $a,b,c$ 。

展开

Using the standard results

\sum_{r=1}^{n}r^2 =\frac{n(n+1)(2n+1)}6

and

\sum_{r=1}^{n}r^3 =\bigg(\frac{n(n+1)}2\bigg)^2,

we have

\begin{align*} \sum_{r=1}^{n}r^2(r+2) =&\,\sum_{r=1}^{n}r^3 +2\sum_{r=1}^{n}r^2\\ =&\,\frac{n^2(n+1)^2}{4}\\ &\,+\frac{n(n+1)(2n+1)}3\\ =&\,\frac1{12}n(n+1)\\ &\,\hspace{2pt}\times \big[3n(n+1)+4(2n+1)\big]\\ =&\,\frac1{12}n(n+1) (3n^2+11n+4). \end{align*}

This is in the required form, so

\boxed{a=3,\qquad b=11,\qquad c=4}.