CalcGospel 國際數學圖譜
返回

FP1 2021 Oct Q5

A Level / Edexcel / FP1

IAL 2021 Oct Paper · Question 5

题目

Problem

5. (a) Use the standard results for ∑r=1nr3\displaystyle\sum_{r=1}^{n} r^3, ∑r=1nr2\displaystyle\sum_{r=1}^{n} r^2 and ∑r=1nr\displaystyle\sum_{r=1}^{n} r to show that for all positive integers nn,

 ∑r=1n[r(r−1) ×(r−3)]= 112n(n+1) ×(n−1) ×(3n−10)\begin{align*} &\,\sum_{r=1}^{n}\big[r(r-1)\\ &\,\hspace{2pt}\times(r-3)\big]\\ =&\,\frac{1}{12}n(n+1)\\ &\,\hspace{2pt}\times(n-1)\\ &\,\hspace{4pt}\times(3n-10) \end{align*}
(5)

(b) Hence show that

 ∑r=n+12n+1[r(r−1) ×(r−3)]= 112n(n+1) ×(an2+bn+c)\begin{align*} &\,\sum_{r=n+1}^{2n+1}\big[r(r-1)\\ &\,\hspace{2pt}\times(r-3)\big]\\ =&\,\frac{1}{12}n(n+1)\\ &\,\hspace{2pt}\times(an^2+bn+c) \end{align*}

where aa, bb and cc are integers to be determined.

(3)
题目中文翻译
  1. (a) 利用 ∑r=1nr3\displaystyle\sum_{r=1}^{n} r^3、∑r=1nr2\displaystyle\sum_{r=1}^{n} r^2 和 ∑r=1nr\displaystyle\sum_{r=1}^{n} r 的标准结果证明，对于所有正整数 nn
 ∑r=1n[r(r−1) ×(r−3)]= 112n(n+1) ×(n−1) ×(3n−10)\begin{align*} &\,\sum_{r=1}^{n}\big[r(r-1)\\ &\,\hspace{2pt}\times(r-3)\big]\\ =&\,\frac{1}{12}n(n+1)\\ &\,\hspace{2pt}\times(n-1)\\ &\,\hspace{4pt}\times(3n-10) \end{align*}
(5)

(b) 由此证明

 ∑r=n+12n+1[r(r−1) ×(r−3)]= 112n(n+1) ×(an2+bn+c)\begin{align*} &\,\sum_{r=n+1}^{2n+1}\big[r(r-1)\\ &\,\hspace{2pt}\times(r-3)\big]\\ =&\,\frac{1}{12}n(n+1)\\ &\,\hspace{2pt}\times(an^2+bn+c) \end{align*}

其中 aa、bb 和 cc 是待定整数。

(3)

解答

(a)

解法一

思路

展开

先展开被求和的代数式，再分别套用三个标准求和公式。提取公因式 112n(n+1)\frac{1}{12}n(n+1) 后，将剩余的二次式因式分解，即可得到题目指定的形式。

答题过程

展开

First,

 r(r−1)(r−3)= r(r2−4r+3)= r3−4r2+3r.\begin{align*} &\,r(r-1)(r-3)\\ =&\,r(r^2-4r+3)\\ =&\,r^3-4r^2+3r. \end{align*}

Using the standard results,

 ∑r=1nr(r−1)(r−3)= ∑r=1nr3− 4∑r=1nr2+ 3∑r=1nr= n2(n+1)24− 4n(n+1)(2n+1)6+ 3n(n+1)2= n2(n+1)24− 2n(n+1)(2n+1)3+ 3n(n+1)2= 112n(n+1) ×[3n(n+1) −8(2n+1)+18].\begin{align*} &\,\sum_{r=1}^{n}r(r-1)(r-3)\\ =&\,\sum_{r=1}^{n}r^3\\ -&\,4\sum_{r=1}^{n}r^2\\ +&\,3\sum_{r=1}^{n}r\\ =&\,\frac{n^2(n+1)^2}{4}\\ -&\,\frac{4n(n+1)(2n+1)}{6}\\ +&\,\frac{3n(n+1)}{2}\\ =&\,\frac{n^2(n+1)^2}{4}\\ -&\,\frac{2n(n+1)(2n+1)}{3}\\ +&\,\frac{3n(n+1)}{2}\\ =&\,\frac{1}{12}n(n+1)\\ &\,\hspace{2pt}\times\big[3n(n+1)\\ &\,\hspace{4pt}-8(2n+1)+18\big]. \end{align*}

The expression in brackets simplifies as follows:

 3n(n+1) −8(2n+1)+18= 3n2+3n−16n −8+18= 3n2−13n+10= 3n2−3n−10n+10= 3n(n−1)−10(n−1)= (n−1)(3n−10).\begin{align*} &\,3n(n+1)\\ &\,-8(2n+1)+18\\ =&\,3n^2+3n-16n\\ &\,-8+18\\ =&\,3n^2-13n+10\\ =&\,3n^2-3n-10n+10\\ =&\,3n(n-1)-10(n-1)\\ =&\,(n-1)(3n-10). \end{align*}

Therefore,

 ∑r=1nr(r−1)(r−3)= 112n(n+1) ×(3n2−13n+10)= 112n(n+1) ×(n−1)(3n−10),\begin{align*} &\,\sum_{r=1}^{n}r(r-1)(r-3)\\ =&\,\frac{1}{12}n(n+1)\\ &\,\hspace{2pt}\times(3n^2-13n+10)\\ =&\,\frac{1}{12}n(n+1)\\ &\,\hspace{2pt}\times(n-1)(3n-10), \end{align*}

as required.

(b)

解法一

思路

展开

承接 (a)，把从 11 到 mm 的和记为 SmS_m。所求区间和等于 S2n+1−SnS_{2n+1}-S_n；代入 (a) 的结果并化简，即可比较系数求出 aa、bb、cc。

答题过程

展开

Let

 Sm= ∑r=1mr(r−1)(r−3).\begin{align*} &\,S_m\\ =&\,\sum_{r=1}^{m}r(r-1)(r-3). \end{align*}

From part (a),

 Sm= 112m(m+1) ×(m−1)(3m−10).\begin{align*} &\,S_m\\ =&\,\frac{1}{12}m(m+1)\\ &\,\hspace{2pt}\times(m-1)(3m-10). \end{align*}

The required sum

∑r=n+12n+1r(r−1)(r−3)\sum_{r=n+1}^{2n+1}r(r-1)(r-3)

is S2n+1−SnS_{2n+1}-S_n. Hence

 S2n+1−Sn= 112(2n+1)(2n+2) ×(2n)(6n−7)− 112n(n+1)(n−1) ×(3n−10).\begin{align*} &\,S_{2n+1}-S_n\\ =&\,\frac{1}{12}(2n+1)(2n+2)\\ &\,\hspace{2pt}\times(2n)(6n-7)\\ -&\,\frac{1}{12}n(n+1)(n-1)\\ &\,\hspace{2pt}\times(3n-10). \end{align*}

Using 2n+2=2(n+1)2n+2=2(n+1), we can factor out 112n(n+1)\frac{1}{12}n(n+1):

 S2n+1−Sn= 112n(n+1) ×[4(2n+1)(6n−7) −(n−1)(3n−10)].\begin{align*} &\,S_{2n+1}-S_n\\ =&\,\frac{1}{12}n(n+1)\\ &\,\hspace{2pt}\times\big[4(2n+1)(6n-7)\\ &\,\hspace{4pt}-(n-1)(3n-10)\big]. \end{align*}

Expanding the expression in brackets,

 4(2n+1)(6n−7) −(n−1)(3n−10)= 4(12n2−8n−7) −(3n2−13n+10)= 48n2−32n−28 −3n2+13n−10= 45n2−19n−38.\begin{align*} &\,4(2n+1)(6n-7)\\ &\,-(n-1)(3n-10)\\ =&\,4(12n^2-8n-7)\\ &\,-(3n^2-13n+10)\\ =&\,48n^2-32n-28\\ &\,-3n^2+13n-10\\ =&\,45n^2-19n-38. \end{align*}

Therefore,

 S2n+1−Sn= 112n(n+1) ×(45n2−19n−38).\begin{align*} &\,S_{2n+1}-S_n\\ =&\,\frac{1}{12}n(n+1)\\ &\,\hspace{2pt}\times(45n^2-19n-38). \end{align*}

Comparing with the required form gives

a= 45,b= −19,c= −38.\boxed{ \begin{aligned} a=&\,45,\\ b=&\,-19,\\ c=&\,-38 \end{aligned} }.