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IAL 2020 Oct Q2

A Level / Edexcel / FP2

IAL 2020 Oct Paper · Question 2

题目

Problem

(a) Write

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

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 \end{align*}

giving your answer in the form

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

分母已经完全分解，所以设成三个简单分式。把两边同乘 r(r−1)(r+1)r(r-1)(r+1) 后比较系数。

#### 答题过程
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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*}

Then

3r+1= A(r−1)(r+1)+Br(r+1)+Cr(r−1)= A(r2−1)+B(r2+r)+C(r2−r)= (A+B+C)r2+(B−C)r−A\begin{align*} 3r+1 =&\,A(r-1)(r+1)+Br(r+1)+Cr(r-1)\\[4mm] =&\,A(r^2-1)+B(r^2+r)+C(r^2-r)\\[4mm] =&\,(A+B+C)r^2+(B-C)r-A \end{align*}

Comparing coefficients,

A+B+C= 0,B−C= 3,−A= 1\begin{align*} A+B+C=&\,0,\\[2mm] B-C=&\,3,\\[2mm] -A=&\,1 \end{align*}

So

A=−1,B=2,C=−1\begin{align*} A=-1,\qquad B=2,\qquad C=-1 \end{align*}

Therefore

3r+1r(r−1)(r+1)= −1r+2r−1−1r+1\begin{align*} \frac{3r+1}{r(r-1)(r+1)} =&\,-\frac1r+\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。为了清楚展示抵消，要写出开头几项和结尾几项，中间项会成对消掉。

答题过程

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Using part (a),

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

Writing out the terms,

Sn= (2−12−13) +(1−13−14) +(23−14−15) +⋯ +(2n−2−1n−1−1n) +(2n−1−1n−1n+1)\begin{align*} S_n =&\,\left(2-\frac12-\frac13\right)\\[2mm] &\,\hspace{2pt}+\left(1-\frac13-\frac14\right)\\[2mm] &\,\hspace{4pt}+\left(\frac23-\frac14-\frac15\right)\\[2mm] &\,\hspace{6pt}+\cdots\\[2mm] &\,\hspace{8pt}+\left(\frac{2}{n-2}-\frac1{n-1}-\frac1n\right)\\[2mm] &\,\hspace{10pt}+\left(\frac{2}{n-1}-\frac1n-\frac1{n+1}\right) \end{align*}

After cancellation,

Sn= 2−12−1n−1n−1n+1= 52−2n−1n+1\begin{align*} S_n =&\,2-\frac12-\frac1n-\frac1n-\frac{1}{n+1}\\[4mm] =&\,\frac52-\frac2n-\frac1{n+1} \end{align*}

Put over the required denominator:

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

(c)

解法一

思路

展开

从 1515 到 2020 的和等于前 2020 项和减去前 1414 项和。这里的“前”是指从本题求和下限 r=2r=2 开始。

答题过程

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Using part (b),

∑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= 13210\begin{align*} \sum_{r=15}^{20}\frac{3r+1}{r(r-1)(r+1)} =&\,S_{20}-S_{14}\\[4mm] =&\,\frac{5(20)^2-20-4}{2(20)(21)} -\frac{5(14)^2-14-4}{2(14)(15)}\\[4mm] =&\,\frac{1976}{840}-\frac{962}{420}\\[4mm] =&\,\frac{247}{105}-\frac{481}{210}\\[4mm] =&\,\frac{13}{210} \end{align*}