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FP2 2026 Jan A Q4

A Level / Edexcel / FP2

IAL 2026 Jan A Paper · Question 4

题目

Problem

4.

(a) Express

4r+2r(r+1)(r+2)\begin{align*} \frac{4r+2}{r(r+1)(r+2)} \end{align*}

in partial fractions.

(3)

(b) Hence, using the method of differences, prove that

∑r=1n4r+2r(r+1)(r+2)= n(an+b)2(n+1)(n+2)\begin{align*} \sum_{r=1}^{n} \frac{4r+2}{r(r+1)(r+2)} =&\, \frac{n(an+b)}{2(n+1)(n+2)} \end{align*}

where aa and bb are constants to be found.

(5)
题目中文翻译

(a) 将

4r+2r(r+1)(r+2)\begin{align*} \frac{4r+2}{r(r+1)(r+2)} \end{align*}

拆成部分分式。

(3)

(b) 由此，使用差分相消法证明

∑r=1n4r+2r(r+1)(r+2)= n(an+b)2(n+1)(n+2)\begin{align*} \sum_{r=1}^{n} \frac{4r+2}{r(r+1)(r+2)} =&\, \frac{n(an+b)}{2(n+1)(n+2)} \end{align*}

其中 aa 和 bb 为待求常数。

(5)

解答

(a)

解法一

思路

展开

分母是三个连续的一次因式，所以设成三项简单分式。代入 r=0,−1,−2r=0,-1,-2 可以很快求出三个常数。

答题过程

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Let

4r+2r(r+1)(r+2)= Ar+Br+1+Cr+2\begin{align*} \frac{4r+2}{r(r+1)(r+2)} =&\, \frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2} \end{align*}

Then

4r+2= A(r+1)(r+2)+Br(r+2)+Cr(r+1)\begin{align*} 4r+2 =&\, 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\Longrightarrow\quad A=1 \end{align*}

Put r=−1r=-1:

−2=−B⟹B=2\begin{align*} -2=-B \quad\Longrightarrow\quad B=2 \end{align*}

Put r=−2r=-2:

−6=2C⟹C=−3\begin{align*} -6=2C \quad\Longrightarrow\quad C=-3 \end{align*}

Therefore

4r+2r(r+1)(r+2)= 1r+2r+1−3r+2\begin{align*} \frac{4r+2}{r(r+1)(r+2)} =&\, \frac1r+\frac{2}{r+1}-\frac{3}{r+2} \end{align*}

(b)

解法一

思路

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把 (a) 的结果逐项展开。中间大部分项会抵消，剩下开头的常数项和结尾含 nn 的项，再通分整理。

答题过程

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Using part (a),

∑r=1n4r+2r(r+1)(r+2)= ∑r=1n(1r+2r+1−3r+2)\begin{align*} \sum_{r=1}^{n} \frac{4r+2}{r(r+1)(r+2)} =&\, \sum_{r=1}^{n} \Big( \frac1r+\frac{2}{r+1}-\frac{3}{r+2} \Big) \end{align*}

Expanding,

∑r=1n(1r+2r+1−3r+2)= (1+22−33) +(12+23−34) +(13+24−35)+⋯ +(1n−1+2n−3n+1) +(1n+2n+1−3n+2)\begin{align*} \sum_{r=1}^{n} \Big( \frac1r+\frac{2}{r+1}-\frac{3}{r+2} \Big) =&\, \Big(1+\frac22-\frac33\Big) \\[2mm] &\,\hspace{2pt}+\Big(\frac12+\frac23-\frac34\Big)\\[2mm] &\,\hspace{4pt}+ \Big(\frac13+\frac24-\frac35\Big) +\cdots\\[3mm] &\,\hspace{6pt}+ \Big(\frac{1}{n-1}+\frac{2}{n}-\frac{3}{n+1}\Big) \\[2mm] &\,\hspace{8pt}+ \Big(\frac{1}{n}+\frac{2}{n+1}-\frac{3}{n+2}\Big) \end{align*}

After cancellation, the initial constant terms and the remaining end terms give:

∑r=1n4r+2r(r+1)(r+2)= 1+22+12−3n+1+2n+1−3n+2= 52−1n+1−3n+2\begin{align*} \sum_{r=1}^{n} \frac{4r+2}{r(r+1)(r+2)} =&\, 1+\frac22+\frac12 -\frac{3}{n+1} +\frac{2}{n+1} -\frac{3}{n+2}\\[3mm] =&\, \frac{5}{2}-\frac{1}{n+1}-\frac{3}{n+2} \end{align*}

Put over a common denominator:

52−1n+1−3n+2= 5(n+1)(n+2)−2(n+2)−6(n+1)2(n+1)(n+2)= 5n2+15n+10−2n−4−6n−62(n+1)(n+2)= 5n2+7n2(n+1)(n+2)= n(5n+7)2(n+1)(n+2)\begin{align*} \frac52-\frac{1}{n+1}-\frac{3}{n+2} =&\, \frac{5(n+1)(n+2)-2(n+2)-6(n+1)} {2(n+1)(n+2)}\\[3mm] =&\, \frac{5n^2+15n+10-2n-4-6n-6} {2(n+1)(n+2)}\\[3mm] =&\, \frac{5n^2+7n}{2(n+1)(n+2)}\\[3mm] =&\, \frac{n(5n+7)}{2(n+1)(n+2)} \end{align*}

Therefore

a=5,b=7\begin{align*} a=5,\qquad b=7 \end{align*}

解法二

思路

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将 (a) 的三个简单分式分成两组相邻差。第一组隔两项相消，第二组逐项相消，分别求和后再合并。

答题过程

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Using the three partial fractions from part (a),

1r+2r+1−3r+2= (1r−1r+2) +2(1r+1−1r+2)\begin{align*} \frac1r+\frac2{r+1}-\frac3{r+2} =&\,\Big(\frac1r-\frac1{r+2}\Big)\\[2mm] &\,\hspace{2pt}+2\Big(\frac1{r+1}-\frac1{r+2}\Big) \end{align*}

Write the two sums as UnU_n and VnV_n. For the first sum,

Un= ∑r=1n(1r−1r+2)= (1−13)+(12−14) +(13−15)+⋯ +(1n−1−1n+1) +(1n−1n+2)= 1+12−1n+1−1n+2\begin{align*} U_n =&\,\sum_{r=1}^{n}\Big(\frac1r-\frac1{r+2}\Big)\\[3mm] =&\,\Big(1-\frac13\Big)+\Big(\frac12-\frac14\Big)\\[2mm] &\,\hspace{2pt}+\Big(\frac13-\frac15\Big)+\cdots\\[2mm] &\,\hspace{4pt}+\Big(\frac1{n-1}-\frac1{n+1}\Big)\\[2mm] &\,\hspace{6pt}+\Big(\frac1n-\frac1{n+2}\Big)\\[3mm] =&\,1+\frac12-\frac1{n+1}-\frac1{n+2} \end{align*}

For the second sum,

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

Hence

∑r=1n4r+2r(r+1)(r+2)= Un+2Vn= 52−1n+1−3n+2= 5(n+1)(n+2)−2(n+2)−6(n+1)2(n+1)(n+2)= 5n2+7n2(n+1)(n+2)= n(5n+7)2(n+1)(n+2)\begin{align*} \sum_{r=1}^{n}\frac{4r+2}{r(r+1)(r+2)} =&\,U_n+2V_n\\[3mm] =&\,\frac52-\frac1{n+1}-\frac3{n+2}\\[3mm] =&\,\frac{5(n+1)(n+2)-2(n+2)-6(n+1)}{2(n+1)(n+2)}\\[3mm] =&\,\frac{5n^2+7n}{2(n+1)(n+2)}\\[3mm] =&\,\frac{n(5n+7)}{2(n+1)(n+2)} \end{align*}

Therefore a=5a=5 and b=7b=7.