题目

Problem

In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

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

\begin{align*} &\,\sum_{r=1}^{n}(7r-5)^2\\ =&\,\frac n6(7n+1)\\ &\,\hspace{2pt}\times(An+B) \end{align*}

where $A$ and $B$ are integers to be determined.

(6)

题目中文翻译

本题必须展示所有解题步骤。 完全依赖计算器技术的解答不可接受。

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

\begin{align*} &\,\sum_{r=1}^{n}(7r-5)^2\\ =&\,\frac n6(7n+1)\\ &\,\hspace{2pt}\times(An+B) \end{align*}

其中 $A$ 和 $B$ 为待确定的整数。

(6)

解答

展开

先展开被求和项，再分别使用 $\sum r^2$ 、 $\sum r$ 与常数和的标准结果。提出公因式 $n/6$ 后，把剩下的式子记为 $T$ ，逐步展开、合并同类项并因式分解，即可确定 $A,B$ 。

展开

First,

\begin{align*} (7r-5)^2=&\,49r^2\\ &\,\hspace{2pt}-70r+25. \end{align*}

Using the standard summation formulae,

\begin{align*} &\,\sum_{r=1}^{n}(7r-5)^2\\ =&\,49\sum_{r=1}^{n}r^2\\ &\,\hspace{2pt}-70\sum_{r=1}^{n}r\\ &\,\hspace{4pt}+\sum_{r=1}^{n}25\\ =&\,49\Big[\frac{n(n+1)(2n+1)}6\Big]\\ &\,\hspace{2pt}-70\Big[\frac{n(n+1)}2\Big]\\ &\,\hspace{4pt}+25n. \end{align*}

Factoring out $n/6$ , write the sum as $\frac n6T$ , where

\begin{align*} T=&\,49(n+1)(2n+1)\\ &\,\hspace{2pt}-210(n+1)+150. \end{align*}

Expanding and collecting terms,

\begin{align*} T=&\,49(2n^2+3n+1)\\ &\,\hspace{2pt}-210n-210+150\\ =&\,98n^2+147n+49\\ &\,\hspace{2pt}-210n-210+150\\ =&\,98n^2-63n-11\\ =&\,98n^2-77n\\ &\,\hspace{2pt}+14n-11\\ =&\,7n(14n-11)\\ &\,\hspace{2pt}+(14n-11)\\ =&\,(7n+1)(14n-11). \end{align*}

Therefore,

\begin{align*} &\,\sum_{r=1}^{n}(7r-5)^2\\ =&\,\frac n6(98n^2-63n-11)\\ =&\,\frac n6(7n+1)\\ &\,\hspace{2pt}\times(14n-11). \end{align*}

Comparing with the required form gives

\boxed{A=14,\qquad B=-11}.

展开

沿用解法一的标准求和步骤，得到提出 $n/6$ 后的二次式。再把题目目标中的 $(7n+1)(An+B)$ 展开，比较各次幂系数，最后将求得的 $A,B$ 代回验证。

展开

Following the summation steps in Method 1,

\begin{align*} &\,\sum_{r=1}^{n}(7r-5)^2\\ =&\,\frac n6(98n^2-63n-11). \end{align*}

Since $n>0$ , comparing this with the required form gives

\begin{align*} &\,(7n+1)(An+B)\\ =&\,98n^2-63n-11. \end{align*}

Expanding the left-hand side,

\begin{align*} &\,(7n+1)(An+B)\\ =&\,7An^2+7Bn+An+B\\ =&\,7An^2+(7B+A)n\\ &\,\hspace{2pt}+B. \end{align*}

Comparing coefficients of $n^2$ , $n$ and the constant term gives

\begin{align*} 7A=&\,98,\\ 7B+A=&\,-63,\\ B=&\,-11. \end{align*}

Thus

\boxed{A=14,\qquad B=-11}.

Checking the coefficient of $n$ ,

\begin{align*} 7B+A=&\,7(-11)+14\\ =&\,-77+14\\ =&\,-63. \end{align*}

Substituting $A=14$ and $B=-11$ back into the product,

\begin{align*} &\,(7n+1)(14n-11)\\ =&\,98n^2-77n+14n-11\\ =&\,98n^2-63n-11. \end{align*}

Hence

\begin{align*} &\,\sum_{r=1}^{n}(7r-5)^2\\ =&\,\frac n6(7n+1)\\ &\,\hspace{2pt}\times(14n-11), \end{align*}

as required.