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IAL 2022 Oct Q2

A Level / Edexcel / FP2

IAL 2022 Oct Paper · Question 2

题目

Problem

(a) Write 3r+1r(r−1)(r+1)\frac{3r+1}{r(r-1)(r+1)} in partial fractions.

(2)

(b) Hence find

∑r=2n3r+1r(r−1)(r+1)n⩾2\begin{align*} \sum_{r=2}^{n} \frac{3r+1}{r(r-1)(r+1)} \qquad n \geqslant 2\\[2mm] \end{align*}

giving your answer in the form

an2+bn+c2n(n+1)\begin{align*} \frac{an^2+bn+c}{2n(n+1)}\\[2mm] \end{align*}

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

(5)

(c) Hence determine the exact value of

∑r=15203r+1r(r−1)(r+1)\begin{align*} \sum_{r=15}^{20} \frac{3r+1}{r(r-1)(r+1)} \end{align*}
(2)

解答

(a)

解法一

思路

展开

分母已经分解成 rr、r−1r-1、r+1r+1 三个一次因式，所以设成三个简单分式，再通分比较系数。

答题过程

展开

Let

3r+1r(r−1)(r+1)= Ar+Br−1+Cr+1.\begin{align*} \frac{3r+1}{r(r-1)(r+1)} =&\,\frac{A}{r}+\frac{B}{r-1}+\frac{C}{r+1}. \end{align*}

Multiplying by r(r−1)(r+1)r(r-1)(r+1),

3r+1= A(r−1)(r+1)+Br(r+1)+Cr(r−1).\begin{align*} 3r+1 =&\,A(r-1)(r+1)+Br(r+1)+Cr(r-1). \end{align*}

Put r=0r=0:

1=−A⟹A=−1.\begin{align*} 1=-A \quad\Longrightarrow\quad A=-1. \end{align*}

Put r=1r=1:

4=2B⟹B=2.\begin{align*} 4=2B \quad\Longrightarrow\quad B=2. \end{align*}

Put r=−1r=-1:

−2=2C⟹C=−1.\begin{align*} -2=2C \quad\Longrightarrow\quad C=-1. \end{align*}

Therefore

3r+1r(r−1)(r+1)=−1r+2r−1−1r+1.\begin{align*} \boxed{ \frac{3r+1}{r(r-1)(r+1)} =-\frac{1}{r}+\frac{2}{r-1}-\frac{1}{r+1} }. \end{align*}

(b)

解法一

思路

展开

把 (a) 的部分分式代入求和。展开前几项和最后几项后，中间大量项会抵消，只剩开头和结尾附近的项。

答题过程

展开

From part (a),

3r+1r(r−1)(r+1)=−1r+2r−1−1r+1.\begin{align*} \frac{3r+1}{r(r-1)(r+1)} =-\frac{1}{r}+\frac{2}{r-1}-\frac{1}{r+1}. \end{align*}

So

∑r=2n3r+1r(r−1)(r+1)= ∑r=2n(−1r+2r−1−1r+1).\begin{align*} \sum_{r=2}^{n} \frac{3r+1}{r(r-1)(r+1)} =&\,\sum_{r=2}^{n} \left(-\frac{1}{r}+\frac{2}{r-1}-\frac{1}{r+1}\right). \end{align*}

Writing out the terms,

 (−12+21−13) +(−13+22−14) +(−14+23−15) +⋯ +(−1n−1+2n−2−1n) +(−1n+2n−1−1n+1).\begin{align*} &\,\left(-\frac{1}{2}+\frac{2}{1}-\frac{1}{3}\right)\\[4mm] &\,\hspace{2pt}+\left(-\frac{1}{3}+\frac{2}{2}-\frac{1}{4}\right)\\[4mm] &\,\hspace{4pt}+\left(-\frac{1}{4}+\frac{2}{3}-\frac{1}{5}\right)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{8pt}+\left(-\frac{1}{n-1}+\frac{2}{n-2}-\frac{1}{n}\right)\\[4mm] &\,\hspace{10pt}+\left(-\frac{1}{n}+\frac{2}{n-1}-\frac{1}{n+1}\right). \end{align*}

After cancellation,

∑r=2n3r+1r(r−1)(r+1)= 2−12+1−2n−1n+1= 52−2n−1n+1.\begin{align*} \sum_{r=2}^{n} \frac{3r+1}{r(r-1)(r+1)} =&\,2-\frac{1}{2}+1-\frac{2}{n}-\frac{1}{n+1}\\[2mm] =&\,\frac{5}{2}-\frac{2}{n}-\frac{1}{n+1}. \end{align*}

Use the common denominator 2n(n+1)2n(n+1):

52−2n−1n+1= 5n(n+1)−4(n+1)−2n2n(n+1)= 5n2+5n−4n−4−2n2n(n+1)= 5n2−n−42n(n+1).\begin{align*} \frac{5}{2}-\frac{2}{n}-\frac{1}{n+1} =&\,\frac{5n(n+1)-4(n+1)-2n}{2n(n+1)}\\[2mm] =&\,\frac{5n^2+5n-4n-4-2n}{2n(n+1)}\\[2mm] =&\,\frac{5n^2-n-4}{2n(n+1)}. \end{align*}

Thus

a=5,b=−1,c=−4.\begin{align*} \boxed{a=5,\qquad b=-1,\qquad c=-4}. \end{align*}

(c)

解法一

思路

展开

由 (b) 的结果可求从 r=2r=2 到 nn 的和。要求 r=15r=15 到 2020，就是

S20−S14.\begin{align*} S_{20}-S_{14}. \end{align*}

答题过程

展开

Let

Sn=∑r=2n3r+1r(r−1)(r+1).\begin{align*} S_n=\sum_{r=2}^{n} \frac{3r+1}{r(r-1)(r+1)}. \end{align*}

From part (b),

Sn=5n2−n−42n(n+1).\begin{align*} S_n=\frac{5n^2-n-4}{2n(n+1)}. \end{align*}

Therefore

∑r=15203r+1r(r−1)(r+1)= S20−S14= 5(20)2−20−42(20)(21) −5(14)2−14−42(14)(15)= 1976840−962420= 247105−481210= 494−481210= 13210.\begin{align*} \sum_{r=15}^{20} \frac{3r+1}{r(r-1)(r+1)} =&\,S_{20}-S_{14}\\[2mm] =&\,\frac{5(20)^2-20-4}{2(20)(21)}\\[2mm] &\,\hspace{2pt}-\frac{5(14)^2-14-4}{2(14)(15)}\\[2mm] =&\,\frac{1976}{840}-\frac{962}{420}\\[2mm] =&\,\frac{247}{105}-\frac{481}{210}\\[2mm] =&\,\frac{494-481}{210}\\[2mm] =&\,\frac{13}{210}. \end{align*}

Hence

13210.\begin{align*} \boxed{\frac{13}{210}}. \end{align*}