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IAL 2023 Jan FP1 Q2

A Level / Edexcel / FP1

IAL 2023 Jan Paper · Question 2

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 ∑r=1nr\displaystyle\sum_{r=1}^{n} r and ∑r=1nr2\displaystyle\sum_{r=1}^{n} r^2 to show that for all positive integers nn

∑r=1n(7r−5)2=n6(7n+1)(An+B)\sum_{r=1}^{n} (7r - 5)^2 = \frac{n}{6}(7n + 1)(An + B)

where AA and BB 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 ∑r=1nr\displaystyle\sum_{r=1}^{n} r 和 ∑r=1nr2\displaystyle\sum_{r=1}^{n} r^2 的标准结果证明：对于所有正整数 nn， ∑r=1n(7r−5)2=n6(7n+1)(An+B)\sum_{r=1}^{n} (7r - 5)^2 = \frac{n}{6}(7n + 1)(An + B)

其中 AA 和 BB 为待确定的整数。

解答

解法一

思路

展开

先展开被求和项，再分别使用 ∑r2\sum r^2、∑r\sum r 与常数和的标准结果。通分并合并同类项后，对所得二次式因式分解，即可与题目指定形式比较出 A,BA,B。

答题过程

展开

First,

(7r−5)2=49r2−70r+25.(7r-5)^2=49r^2-70r+25.

Therefore,

∑r=1n(7r−5)2= 49∑r=1nr2−70∑r=1nr+∑r=1n25= 49[n(n+1)(2n+1)6] −70[n(n+1)2]+25n= n6[49(n+1)(2n+1) −210(n+1)+150]= n6(98n2−63n−11)= n6(7n+1)(14n−11).\begin{align*} \sum_{r=1}^{n}(7r-5)^2 =&\,49\sum_{r=1}^{n}r^2 -70\sum_{r=1}^{n}r +\sum_{r=1}^{n}25\\ =&\,49\left[ \frac{n(n+1)(2n+1)}6 \right]\\ &\,-70\left[\frac{n(n+1)}2\right] +25n\\ =&\,\frac n6\big[ 49(n+1)(2n+1)\\ &\,\hspace{8pt}-210(n+1)+150 \big]\\ =&\,\frac n6(98n^2-63n-11)\\ =&\,\frac n6(7n+1)(14n-11). \end{align*}

Hence

A=14,B=−11.\boxed{A=14,\qquad B=-11}.

解法二

思路

展开

官方替代路线是在得到展开后的二次式后，不直接因式分解，而把题目目标中的 (7n+1)(An+B)(7n+1)(An+B) 展开并比较各次幂系数。这也能系统地确定 A,BA,B。

答题过程

展开

Following the summation steps in Method 1,

∑r=1n(7r−5)2=n6(98n2−63n−11).\sum_{r=1}^{n}(7r-5)^2 =\frac n6(98n^2-63n-11).

On the other hand,

(7n+1)(An+B)= 7An2+(7B+A)n+B.\begin{align*} (7n+1)(An+B) =&\,7An^2+(7B+A)n+B. \end{align*}

Comparing this with

98n2−63n−11,98n^2-63n-11,

gives

7A=98⟹A=14,7A=98 \quad\Longrightarrow\quad A=14,

and

B=−11.B=-11.

The coefficient of nn is then

7B+A=7(−11)+14=−63,7B+A=7(-11)+14=-63,

which verifies the comparison. Thus

A=14,B=−11,\boxed{A=14,\qquad B=-11},

and hence

∑r=1n(7r−5)2=n6(7n+1)(14n−11).\boxed{ \sum_{r=1}^{n}(7r-5)^2 =\frac n6(7n+1)(14n-11) }.