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FP2 2025 June Q5

A Level / Edexcel / FP2

IAL 2025 June Paper · Question 5

题目

Problem

5.

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

(a) Express

2r(r+1)(r+2)\begin{align*} \frac{2}{r(r+1)(r+2)} \end{align*}

in partial fractions.

(2)

(b) Use the answer to part (a) and the method of differences to show that

∑r=1n2r(r+1)(r+2)= n(n+a)2(n+b)(n+c)\begin{align*} \sum_{r=1}^{n} \frac{2}{r(r+1)(r+2)} =&\, \frac{n(n+a)}{2(n+b)(n+c)} \end{align*}

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

(5)

(c) Hence form and solve a quadratic inequality to determine the smallest value of nn for which

∑r=1n2r(r+1)(r+2)>715\begin{align*} \sum_{r=1}^{n} \frac{2}{r(r+1)(r+2)} > \frac7{15} \end{align*}
(3)
题目中文翻译

本题必须展示所有计算步骤。

完全依赖计算器技术得出的解答不予接受。

(a) 将

2r(r+1)(r+2)\begin{align*} \frac2{r(r+1)(r+2)} \end{align*}

分解为部分分式。

(2)

(b) 利用 (a) 的结果和逐项相消法，证明

∑r=1n2r(r+1)(r+2)= n(n+a)2(n+b)(n+c),\begin{align*} \sum_{r=1}^n\frac2{r(r+1)(r+2)} =&\,\frac{n(n+a)}{2(n+b)(n+c)}, \end{align*}

并求出整数 aa、bb、cc 的值。

(5)

(c) 由此列出并解一个二次不等式，求满足

∑r=1n2r(r+1)(r+2)> 715\begin{align*} \sum_{r=1}^n\frac2{r(r+1)(r+2)} >&\,\frac7{15} \end{align*}

的最小 nn 值。

(3)

解答

(a)

解法一

思路

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分母有三个不同的一次因式，所以设成三个简单分式。用 r=0,−1,−2r=0,-1,-2 可以快速求系数。

答题过程

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Let

2r(r+1)(r+2)≡ Ar+Br+1+Cr+2\begin{align*} \frac{2}{r(r+1)(r+2)} \equiv&\, \frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2} \end{align*}

Multiply by r(r+1)(r+2)r(r+1)(r+2):

2≡ A(r+1)(r+2)+Br(r+2)+Cr(r+1)\begin{align*} 2 \equiv&\, A(r+1)(r+2)+Br(r+2)+Cr(r+1) \end{align*}

Put r=0r=0:

2=2A⇒A=1\begin{align*} 2=2A \quad \Rightarrow \quad A=1 \end{align*}

Put r=−1r=-1:

2=−B⇒B=−2\begin{align*} 2=-B \quad \Rightarrow \quad B=-2 \end{align*}

Put r=−2r=-2:

2=2C⇒C=1\begin{align*} 2=2C \quad \Rightarrow \quad C=1 \end{align*}

Therefore

2r(r+1)(r+2)=1r−2r+1+1r+2\begin{align*} \boxed{ \frac{2}{r(r+1)(r+2)} = \frac1r-\frac2{r+1}+\frac1{r+2} } \end{align*}

(b)

解法一

思路

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把部分分式逐项展开。由于每一行是 1r−2r+1+1r+2\frac1r-\frac2{r+1}+\frac1{r+2}，中间项会抵消，最后留下开头和结尾附近的几项。

答题过程

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Using part (a),

Sn= ∑r=1n(1r−2r+1+1r+2)\begin{align*} S_n =&\, \sum_{r=1}^{n} \bigg( \frac1r-\frac2{r+1}+\frac1{r+2} \bigg) \end{align*}

Expand the terms for r=1,2,3,…,n−1,nr=1,2,3,\dots,n-1,n:

Sn= (1−22+13) +(12−23+14) +(13−24+15)+⋯ +(1n−1−2n+1n+1) +(1n−2n+1+1n+2).\begin{align*} S_n=&\,\bigg(1-\frac22+\frac13\bigg)\\[3mm] &\,\hspace{2pt}+\bigg(\frac12-\frac23+\frac14\bigg)\\[3mm] &\,\hspace{4pt}+\bigg(\frac13-\frac24+\frac15\bigg)+\cdots\\[3mm] &\,\hspace{6pt}+\bigg(\frac1{n-1}-\frac2n+\frac1{n+1}\bigg)\\[3mm] &\,\hspace{8pt}+\bigg(\frac1n-\frac2{n+1}+\frac1{n+2}\bigg). \end{align*}

To find the sum, we group terms by their denominators:

  • Denominator 11: 11 (from r=1r=1)
  • Denominator 22: −22+12=−1+12=−12-\frac{2}{2} + \frac{1}{2} = -1 + \frac{1}{2} = -\frac{1}{2} (from r=1,2r=1, 2)
  • Denominator kk (for 3⩽k⩽n3 \leqslant k \leqslant n): 1k−2k+1k=0\frac{1}{k} - \frac{2}{k} + \frac{1}{k} = 0 (from r=k−2,k−1,kr=k-2, k-1, k)
  • Denominator n+1n+1: −2n+1+1n+1=−1n+1-\frac{2}{n+1} + \frac{1}{n+1} = -\frac{1}{n+1} (from r=n−1,nr=n-1, n)
  • Denominator n+2n+2: 1n+2\frac{1}{n+2} (from r=nr=n)

Therefore, all terms with denominators from 33 to nn cancel out completely, and the sum simplifies to:

Sn= 1−12−1n+1+1n+2= 12−1n+1+1n+2\begin{align*} S_n =&\, 1 - \frac{1}{2} - \frac{1}{n+1} + \frac{1}{n+2}\\[3mm] =&\, \frac{1}{2}-\frac{1}{n+1}+\frac{1}{n+2} \end{align*}

Put over a common denominator:

Sn= (n+1)(n+2)−2(n+2)+2(n+1)2(n+1)(n+2)= n2+3n+2−2n−4+2n+22(n+1)(n+2)= n2+3n2(n+1)(n+2)= n(n+3)2(n+1)(n+2)\begin{align*} S_n =&\, \frac{ (n+1)(n+2)-2(n+2)+2(n+1) } {2(n+1)(n+2)}\\[3mm] =&\, \frac{ n^2+3n+2-2n-4+2n+2 } {2(n+1)(n+2)}\\[3mm] =&\, \frac{n^2+3n}{2(n+1)(n+2)}\\[3mm] =&\, \frac{n(n+3)}{2(n+1)(n+2)} \end{align*}

Hence

a=3,b=1,c=2\begin{align*} \boxed{a=3,\quad b=1,\quad c=2} \end{align*}

解法二

思路

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也可以把部分分式看成两组相邻差：

1r−2r+1+1r+2=(1r−1r+1)+(−1r+1+1r+2)\begin{align*} \frac1r-\frac2{r+1}+\frac1{r+2} = \bigg(\frac1r-\frac1{r+1}\bigg) +\bigg(-\frac1{r+1}+\frac1{r+2}\bigg) \end{align*}

这样每组都是标准望远镜级数。

答题过程

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From part (a),

1r−2r+1+1r+2= (1r−1r+1)+(−1r+1+1r+2)\begin{align*} \frac1r-\frac2{r+1}+\frac1{r+2} =&\, \bigg(\frac1r-\frac1{r+1}\bigg) +\bigg(-\frac1{r+1}+\frac1{r+2}\bigg) \end{align*}

The first sum telescopes as follows:

∑r=1n(1r−1r+1)= (1−12) +(12−13) +(13−14)+⋯ +(1n−1−1n) +(1n−1n+1)= 1−1n+1.\begin{align*} \sum_{r=1}^n\bigg(\frac1r-\frac1{r+1}\bigg) =&\,\bigg(1-\frac12\bigg)\\[3mm] &\,\hspace{2pt}+\bigg(\frac12-\frac13\bigg)\\[3mm] &\,\hspace{4pt}+\bigg(\frac13-\frac14\bigg)+\cdots\\[3mm] &\,\hspace{6pt}+\bigg(\frac1{n-1}-\frac1n\bigg)\\[3mm] &\,\hspace{8pt}+\bigg(\frac1n-\frac1{n+1}\bigg)\\[3mm] =&\,1-\frac1{n+1}. \end{align*}

Similarly,

∑r=1n(−1r+1+1r+2)= (−12+13) +(−13+14) +(−14+15)+⋯ +(−1n+1n+1) +(−1n+1+1n+2)= −12+1n+2.\begin{align*} \sum_{r=1}^n\bigg(-\frac1{r+1}+\frac1{r+2}\bigg) =&\,\bigg(-\frac12+\frac13\bigg)\\[3mm] &\,\hspace{2pt}+\bigg(-\frac13+\frac14\bigg)\\[3mm] &\,\hspace{4pt}+\bigg(-\frac14+\frac15\bigg)+\cdots\\[3mm] &\,\hspace{6pt}+\bigg(-\frac1n+\frac1{n+1}\bigg)\\[3mm] &\,\hspace{8pt}+\bigg(-\frac1{n+1}+\frac1{n+2}\bigg)\\[3mm] =&\,-\frac12+\frac1{n+2}. \end{align*}

Therefore

Sn= ∑r=1n(1r−1r+1)+∑r=1n(−1r+1+1r+2)= (1−1n+1)+(−12+1n+2)= 12−1n+1+1n+2= n(n+3)2(n+1)(n+2)\begin{align*} S_n =&\, \sum_{r=1}^{n} \bigg(\frac1r-\frac1{r+1}\bigg) +\sum_{r=1}^{n} \bigg(-\frac1{r+1}+\frac1{r+2}\bigg)\\[3mm] =&\, \bigg(1-\frac1{n+1}\bigg) +\bigg(-\frac12+\frac1{n+2}\bigg)\\[3mm] =&\, \frac12-\frac1{n+1}+\frac1{n+2}\\[3mm] =&\, \frac{n(n+3)}{2(n+1)(n+2)} \end{align*}

Hence

a=3,b=1,c=2\begin{align*} \boxed{a=3,\quad b=1,\quad c=2} \end{align*}

解法三

思路

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将 (a) 的部分分式重新组合成两个相邻乘积分母的差。这样相邻两行会直接抵消，只留下首项和末项，求和会更短。

答题过程

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Regroup the partial fractions from part (a):

1r−2r+1+1r+2= (1r−1r+1) −(1r+1−1r+2)= 1r(r+1)−1(r+1)(r+2).\begin{align*} \frac1r-\frac2{r+1}+\frac1{r+2} =&\,\bigg(\frac1r-\frac1{r+1}\bigg)\\[3mm] &\,\hspace{2pt}-\bigg(\frac1{r+1}-\frac1{r+2}\bigg)\\[3mm] =&\,\frac1{r(r+1)}-\frac1{(r+1)(r+2)}. \end{align*}

Hence

Sn= (11⋅2−12⋅3) +(12⋅3−13⋅4) +(13⋅4−14⋅5)+⋯ +(1(n−1)n−1n(n+1)) +(1n(n+1)−1(n+1)(n+2)).\begin{align*} S_n=&\,\bigg(\frac1{1\cdot2}-\frac1{2\cdot3}\bigg)\\[3mm] &\,\hspace{2pt}+\bigg(\frac1{2\cdot3}-\frac1{3\cdot4}\bigg)\\[3mm] &\,\hspace{4pt}+\bigg(\frac1{3\cdot4}-\frac1{4\cdot5}\bigg)+\cdots\\[3mm] &\,\hspace{6pt}+\bigg(\frac1{(n-1)n}-\frac1{n(n+1)}\bigg)\\[3mm] &\,\hspace{8pt}+\bigg(\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\bigg). \end{align*}

After cancellation,

Sn= 12−1(n+1)(n+2)= (n+1)(n+2)−22(n+1)(n+2)= n(n+3)2(n+1)(n+2).\begin{align*} S_n=&\,\frac12-\frac1{(n+1)(n+2)}\\[3mm] =&\,\frac{(n+1)(n+2)-2}{2(n+1)(n+2)}\\[3mm] =&\,\frac{n(n+3)}{2(n+1)(n+2)}. \end{align*}

Thus a=3a=3, b=1b=1 and c=2c=2.

(c)

解法一

思路

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用 (b) 的结果建立不等式。因为 nn 是正整数，且 2(n+1)(n+2)>02(n+1)(n+2)>0，所以可以直接交叉相乘，最后解二次不等式并取最小正整数。

答题过程

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Using part (b), we need

n(n+3)2(n+1)(n+2)>715\begin{align*} \frac{n(n+3)}{2(n+1)(n+2)} >& \frac7{15} \end{align*}

Since nn is positive, 2(n+1)(n+2)>02(n+1)(n+2)>0. Cross-multiply:

15n(n+3)>14(n+1)(n+2)15n2+45n>14(n2+3n+2)15n2+45n>14n2+42n+28n2+3n−28>0\begin{align*} 15n(n+3) >& 14(n+1)(n+2)\\[2mm] 15n^2+45n >& 14(n^2+3n+2)\\[2mm] 15n^2+45n >& 14n^2+42n+28\\[2mm] n^2+3n-28 >& 0 \end{align*}

Factorise:

(n+7)(n−4)>0\begin{align*} (n+7)(n-4)>0 \end{align*}

Thus

n<−7orn>4\begin{align*} n<-7 \quad \text{or} \quad n>4 \end{align*}

Since nn is a positive integer, the smallest possible value is

n=5\begin{align*} \boxed{n=5} \end{align*}