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CIE 9231 2023 June Paper 13 Q2

A Level / CIE / FP1

CIE 9231 2023 June Paper 13 Paper · Question 2

题目

Problem

(a) Use standard results from the list of formulae (MF19) to show that

∑r=1n(6r2+6r−5)=an3+bn2+cn,\sum_{r=1}^{n}(6r^2+6r-5)=an^3+bn^2+cn,

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

(2)

(b) Use the method of differences to find ∑r=1n6r2+6r−5r2+r\displaystyle\sum_{r=1}^{n}\frac{6r^2+6r-5}{r^2+r} in terms of nn.

(4)

(c) Find also ∑r=n+12n6r2+6r−5r2+r\displaystyle\sum_{r=n+1}^{2n}\frac{6r^2+6r-5}{r^2+r} in terms of nn.

(2)
题目中文翻译

(a) 利用公式表（MF19）中的标准结果，证明

∑r=1n(6r2+6r−5)=an3+bn2+cn,\sum_{r=1}^{n}(6r^2+6r-5)=an^3+bn^2+cn,

其中 aa、bb 和 cc 是待求的整数。

(2)

(b) 用差分法求 ∑r=1n6r2+6r−5r2+r\displaystyle\sum_{r=1}^{n}\frac{6r^2+6r-5}{r^2+r}，用 nn 表示。

(4)

(c) 另求 ∑r=n+12n6r2+6r−5r2+r\displaystyle\sum_{r=n+1}^{2n}\frac{6r^2+6r-5}{r^2+r}，用 nn 表示。

(2)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

逐项代入平方和与自然数和，最后展开读出三个整数系数。

答题过程

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Using the standard formulae,

∑r=1n(6r2+6r−5)= n(n+1)(2n+1) +3n(n+1)−5n= 2n3+3n2+n+3n2+3n−5n= 2n3+6n2−n.\begin{align*} \sum_{r=1}^n(6r^2+6r-5) =&\,n(n+1)(2n+1)\\[4mm] &\,+3n(n+1)-5n\\[4mm] =&\,2n^3+3n^2+n+3n^2+3n-5n\\[4mm] =&\,2n^3+6n^2-n. \end{align*}

Therefore a=2,b=6,c=−1\boxed{a=2,\quad b=6,\quad c=-1}.

(b)

解法一

思路

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先做除法分出常数6，剩下的倒数乘积用部分分式形成相邻差。

答题过程

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Division and partial fractions give

6r2+6r−5r2+r=6−5r(r+1),\frac{6r^2+6r-5}{r^2+r} =6-\frac5{r(r+1)},

and

1r(r+1)=1r−1r+1.\frac1{r(r+1)}=\frac1r-\frac1{r+1}.

Thus

∑r=1n1r(r+1)= (1−12) +(12−13) +(13−14)+⋯ +(1n−2−1n−1) +(1n−1−1n) +(1n−1n+1)= 1−1n+1.\begin{align*} \sum_{r=1}^n\frac1{r(r+1)} =&\,\bigg(1-\frac12\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac12-\frac13\bigg)\\[4mm] &\,\hspace{4pt}+\bigg(\frac13-\frac14\bigg)+\cdots\\[4mm] &\,\hspace{6pt}+\bigg(\frac1{n-2}-\frac1{n-1}\bigg)\\[4mm] &\,\hspace{8pt}+\bigg(\frac1{n-1}-\frac1n\bigg)\\[4mm] &\,\hspace{4pt}+\bigg(\frac1n-\frac1{n+1}\bigg)\\[4mm] =&\,1-\frac1{n+1}. \end{align*}

The expanded pattern applies for sufficiently large nn; the final identity also holds directly for smaller positive nn.

Therefore

∑r=1n6r2+6r−5r2+r= 6n−5(1−1n+1)= 6n−5+5n+1.\begin{align*} \sum_{r=1}^n\frac{6r^2+6r-5}{r^2+r} =&\,6n-5\bigg(1-\frac1{n+1}\bigg)\\[4mm] =&\,\boxed{6n-5+\frac5{n+1}}. \end{align*}

(c)

解法一

思路

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从n+1到2n的和，是前2n项减前n项；两个常数−5抵消。

答题过程

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Let TnT_n be the sum in (b). Then

 ∑r=n+12n6r2+6r−5r2+r= T2n−Tn= 12n−5+52n+1 −(6n−5+5n+1)= 6n+52n+1−5n+1= 6n−5n(n+1)(2n+1).\begin{align*} &\,\sum_{r=n+1}^{2n}\frac{6r^2+6r-5}{r^2+r}\\[4mm] =&\,T_{2n}-T_n\\[4mm] =&\,12n-5+\frac5{2n+1}\\[4mm] &\,-\bigg(6n-5+\frac5{n+1}\bigg)\\[4mm] =&\,6n+\frac5{2n+1}-\frac5{n+1}\\[4mm] =&\,\boxed{6n-\frac{5n}{(n+1)(2n+1)}}. \end{align*}