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FP1 2024 Jan Q8

A Level / Edexcel / FP1

IAL 2024 Jan Paper · Question 8

题目

Problem

(a) Use the standard results for summations to show that, for all positive integers nn,

 ∑r=1nr(2r2−3r−1)= 12n(n+1)2(n−2)\begin{align*} &\,\sum_{r=1}^{n}r(2r^2-3r-1)\\ =&\,\frac12n(n+1)^2(n-2) \end{align*}
(4)

(b) Hence show that, for all positive integers nn,

 ∑r=n2nr(2r2−3r−1)= 12n(n−1) ×(an+b)(cn+d)\begin{align*} &\,\sum_{r=n}^{2n}r(2r^2-3r-1)\\ =&\,\frac12n(n-1)\\ &\,\hspace{2pt}\times(an+b)(cn+d) \end{align*}

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

(4)
题目中文翻译

(a) 使用求和的标准结果证明：对于所有正整数 nn，

 ∑r=1nr(2r2−3r−1)= 12n(n+1)2(n−2)\begin{align*} &\,\sum_{r=1}^{n}r(2r^2-3r-1)\\ =&\,\frac12n(n+1)^2(n-2) \end{align*}
(4)

(b) 由此证明：对于所有正整数 nn，

 ∑r=n2nr(2r2−3r−1)= 12n(n−1) ×(an+b)(cn+d)\begin{align*} &\,\sum_{r=n}^{2n}r(2r^2-3r-1)\\ =&\,\frac12n(n-1)\\ &\,\hspace{2pt}\times(an+b)(cn+d) \end{align*}

其中 aa、bb、cc、dd 为待确定的整数。

(4)

解答

(a)

解法一

思路

展开

先展开被求和项，再分别使用 ∑r\sum r、∑r2\sum r^2 与 ∑r3\sum r^3 的标准结果。提出公因式 12n(n+1)\frac12n(n+1) 后，单独整理剩下的二次式，再因式分解。

答题过程

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Let

Sn=∑r=1nr(2r2−3r−1).S_n=\sum_{r=1}^{n}r(2r^2-3r-1).

Using the standard results for summations,

 Sn= 2∑r=1nr3 −3∑r=1nr2 −∑r=1nr= 2[n2(n+1)24] −3[n(n+1)(2n+1)6] −n(n+1)2= 12n2(n+1)2 −12n(n+1)(2n+1) −12n(n+1).\begin{align*} &\,S_n\\ =&\,2\sum_{r=1}^{n}r^3\\ &\,\hspace{2pt}-3\sum_{r=1}^{n}r^2\\ &\,\hspace{4pt}-\sum_{r=1}^{n}r\\ =&\,2\Big[\frac{n^2(n+1)^2}{4}\Big]\\ &\,\hspace{2pt}-3\Big[\frac{n(n+1)(2n+1)}6\Big]\\ &\,\hspace{4pt}-\frac{n(n+1)}2\\ =&\,\frac12n^2(n+1)^2\\ &\,\hspace{2pt}-\frac12n(n+1)(2n+1)\\ &\,\hspace{4pt}-\frac12n(n+1). \end{align*}

Factorising out 12n(n+1)\frac12n(n+1) gives

Sn=12n(n+1)B,S_n=\frac12n(n+1)B,

where

 B= n(n+1)−(2n+1)−1= n2+n−2n−1−1= n2−n−2= (n−2)(n+1).\begin{align*} &\,B\\ =&\,n(n+1)-(2n+1)-1\\ =&\,n^2+n-2n-1-1\\ =&\,n^2-n-2\\ =&\,(n-2)(n+1). \end{align*}

Therefore,

 Sn= 12n(n+1)(n2−n−2)= 12n(n+1)(n−2) ×(n+1)= 12n(n+1)2(n−2).\begin{align*} &\,S_n\\ =&\,\frac12n(n+1)(n^2-n-2)\\ =&\,\frac12n(n+1)(n-2)\\ &\,\hspace{2pt}\times(n+1)\\ =&\,\boxed{\frac12n(n+1)^2(n-2)}. \end{align*}

(b)

解法一

思路

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设 (a) 的部分和为 F(n)F(n)。由于本题从 r=nr=n 开始且包含第 nn 项，应计算 F(2n)−F(n−1)F(2n)-F(n-1)。代入后提出公因式 12n(n−1)\frac12n(n-1)，单独展开括号内的式子，再进行因式分解。

答题过程

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From part (a), let

 F(n)= ∑r=1nr(2r2−3r−1)= 12n(n+1)2(n−2).\begin{align*} &\,F(n)\\ =&\,\sum_{r=1}^{n}r(2r^2-3r-1)\\ =&\,\frac12n(n+1)^2(n-2). \end{align*}

In particular,

 F(2n)= 12(2n)(2n+1)2(2n−2)\begin{align*} &\,F(2n)\\ =&\,\frac12(2n)(2n+1)^2(2n-2) \end{align*}

and

 F(n−1)= 12(n−1)n2(n−3).\begin{align*} &\,F(n-1)\\ =&\,\frac12(n-1)n^2(n-3). \end{align*}

The required sum is

T=∑r=n2nr(2r2−3r−1).T=\sum_{r=n}^{2n}r(2r^2-3r-1).

Since the term with r=nr=n is included,

 T= F(2n)−F(n−1)= 12(2n)(2n+1)2(2n−2) −12(n−1)n2(n−3).\begin{align*} &\,T\\ =&\,F(2n)-F(n-1)\\ =&\,\frac12(2n)(2n+1)^2(2n-2)\\ &\,\hspace{2pt}-\frac12(n-1)n^2(n-3). \end{align*}

Using 2n−2=2(n−1)2n-2=2(n-1) and factorising out 12n(n−1)\frac12n(n-1) gives

T=12n(n−1)C,T=\frac12n(n-1)C,

where

 C= 4(2n+1)2−n(n−3)= 4(4n2+4n+1) −n2+3n= 16n2+16n+4 −n2+3n= 15n2+19n+4= 15n2+15n+4n+4= 15n(n+1)+4(n+1)= (15n+4)(n+1).\begin{align*} &\,C\\ =&\,4(2n+1)^2-n(n-3)\\ =&\,4(4n^2+4n+1)\\ &\,\hspace{2pt}-n^2+3n\\ =&\,16n^2+16n+4\\ &\,\hspace{2pt}-n^2+3n\\ =&\,15n^2+19n+4\\ =&\,15n^2+15n+4n+4\\ =&\,15n(n+1)+4(n+1)\\ =&\,(15n+4)(n+1). \end{align*}

Hence,

 T= 12n(n−1) ×(15n2+19n+4)= 12n(n−1)(15n+4) ×(n+1).\begin{align*} &\,T\\ =&\,\frac12n(n-1)\\ &\,\hspace{2pt}\times(15n^2+19n+4)\\ =&\,\frac12n(n-1)(15n+4)\\ &\,\hspace{2pt}\times(n+1). \end{align*}

Therefore, one valid set of integer values is

a= 15,b= 4,c= 1,d= 1.\begin{align*} a=&\,15,& b=&\,4,\\ c=&\,1,& d=&\,1. \end{align*}