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FP2 2026 Jan Q1

A Level / Edexcel / FP2

IAL 2026 Jan Paper · Question 1

题目

Problem

1.

(a) Express

1(3n−1)(3n+5)\begin{align*} \frac{1}{(3n-1)(3n+5)} \end{align*}

in partial fractions.

(2)

(b) Hence, using the method of differences, show that for all positive integer values of nn,

∑r=1n20(3r−1)(3r+5)≡n(An+B)(3n+C)(3n+D)\begin{align*} \sum_{r=1}^{n}\frac{20}{(3r-1)(3r+5)} \equiv \frac{n(An+B)}{(3n+C)(3n+D)} \end{align*}

where AA, BB, CC and DD are integers to be determined.

(4)
题目中文翻译

(a) 将

1(3n−1)(3n+5)\begin{align*} \frac1{(3n-1)(3n+5)} \end{align*}

分解为部分分式。

(2)

(b) 由此利用差分相消法，证明对于所有正整数 nn，

∑r=1n20(3r−1)(3r+5)≡ n(An+B)(3n+C)(3n+D)\begin{align*} \sum_{r=1}^{n}\frac{20}{(3r-1)(3r+5)} \equiv&\,\frac{n(An+B)}{(3n+C)(3n+D)} \end{align*}

并求出整数 AA、BB、CC 和 DD。

(4)

解答

(a)

解法一

思路

展开

两个一次因式相乘，所以设成两个简单分式。代入方便的 nn 值可以快速求出常数。

答题过程

展开

Let

1(3n−1)(3n+5)= A3n−1+B3n+5\begin{align*} \frac{1}{(3n-1)(3n+5)} =&\, \frac{A}{3n-1}+\frac{B}{3n+5} \end{align*}

Then

1= A(3n+5)+B(3n−1)\begin{align*} 1 =&\,A(3n+5)+B(3n-1) \end{align*}

When 3n−1=03n-1=0, n=13n=\frac13, so

1=6A⟹A=16\begin{align*} 1=6A \quad\Longrightarrow\quad A=\frac16 \end{align*}

When 3n+5=03n+5=0, n=−53n=-\frac53, so

1=−6B⟹B=−16\begin{align*} 1=-6B \quad\Longrightarrow\quad B=-\frac16 \end{align*}

Therefore

1(3n−1)(3n+5)= 16(3n−1)−16(3n+5)\begin{align*} \frac{1}{(3n-1)(3n+5)} =&\, \frac{1}{6(3n-1)} - \frac{1}{6(3n+5)} \end{align*}

(b)

解法一

思路

展开

把 (a) 的结果乘以 2020 后求和。由于分母相差 66，展开时中间项会隔项抵消，最后只剩开头两个正项和结尾两个负项。

答题过程

展开

Using part (a),

20(3r−1)(3r+5)= 206(13r−1−13r+5)\begin{align*} \frac{20}{(3r-1)(3r+5)} =&\, \frac{20}{6} \Big( \frac{1}{3r-1} - \frac{1}{3r+5} \Big) \end{align*}

Hence

∑r=1n20(3r−1)(3r+5)= 206∑r=1n(13r−1−13r+5)\begin{align*} \sum_{r=1}^{n} \frac{20}{(3r-1)(3r+5)} =&\, \frac{20}{6} \sum_{r=1}^{n} \Big( \frac{1}{3r-1} - \frac{1}{3r+5} \Big) \end{align*}

Expanding the sum,

206[(12−18)+(15−111) +(18−114)+⋯ +(13n−7−13n−1) +(13n−4−13n+2) +(13n−1−13n+5)]\begin{align*} {}& \frac{20}{6} \bigg[ \Big(\frac12-\frac18\Big) + \Big(\frac15-\frac{1}{11}\Big)\\[3mm] &\,\hspace{2pt}+ \Big(\frac18-\frac{1}{14}\Big) +\cdots\\[3mm] &\,\hspace{4pt}+ \Big(\frac{1}{3n-7}-\frac{1}{3n-1}\Big)\\[3mm] &\,\hspace{6pt}+ \Big(\frac{1}{3n-4}-\frac{1}{3n+2}\Big)\\[3mm] &\,\hspace{8pt}+ \Big(\frac{1}{3n-1}-\frac{1}{3n+5}\Big) \bigg] \end{align*}

So all the middle terms cancel and

∑r=1n20(3r−1)(3r+5)= 206(12+15−13n+2−13n+5)= 206(710−6n+7(3n+2)(3n+5))= 206(7(3n+2)(3n+5)−10(6n+7)10(3n+2)(3n+5))= 206(63n2+87n10(3n+2)(3n+5))= n(21n+29)(3n+2)(3n+5)\begin{align*} \sum_{r=1}^{n} \frac{20}{(3r-1)(3r+5)} =&\, \frac{20}{6} \Big( \frac12+\frac15 - \frac{1}{3n+2} - \frac{1}{3n+5} \Big)\\[3mm] =&\, \frac{20}{6} \Big( \frac{7}{10} - \frac{6n+7}{(3n+2)(3n+5)} \Big)\\[3mm] =&\, \frac{20}{6} \Big( \frac{7(3n+2)(3n+5)-10(6n+7)} {10(3n+2)(3n+5)} \Big)\\[3mm] =&\, \frac{20}{6} \Big( \frac{63n^2+87n}{10(3n+2)(3n+5)} \Big)\\[3mm] =&\, \frac{n(21n+29)}{(3n+2)(3n+5)} \end{align*}

Therefore

A=21,B=29,C=2,D=5\begin{align*} A=21,\qquad B=29,\qquad C=2,\qquad D=5 \end{align*}