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FP2 2020 Oct Q2

A Level / Edexcel / FP2

IAL 2020 Oct Paper · Question 2

题目

Problem

2.

(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)}\\[3mm] &\,n\geqslant2 \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)
题目中文翻译

(a) 将

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

分解为部分分式。

(2)

(b) 由此求

 ∑r=2n3r+1r(r−1)(r+1) n⩾2\begin{align*} &\,\sum_{r=2}^{n}\frac{3r+1}{r(r-1)(r+1)}\\[3mm] &\,n\geqslant2 \end{align*}

将答案写成

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

的形式，其中 aa、bb、cc 是待求的整数。

(5)

(c) 由此求下列和的精确值：

∑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

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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

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

Write

F(r)= Ar+Br−1 +Cr+1.\begin{align*} F(r)=&\,\frac Ar+\frac B{r-1}\\[2mm] &\,+\frac C{r+1}. \end{align*}

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

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

Hence

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

so

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

(b)

解法一

思路

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把 (a) 的部分分式代入求和，写出前三项和末两项。抵消后，前端留下 2−12+222-\frac12+\frac22；不能漏掉来自第二项的 22=1\frac22=1。

答题过程

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Using part (a), let

Sn= ∑r=2nF(r).\begin{align*} S_n=&\,\sum_{r=2}^n F(r). \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=&\,\big(2-\tfrac12-\tfrac13\big)\\[2mm] &\,+\big(1-\tfrac13-\tfrac14\big)\\[2mm] &\,+\big(\tfrac23-\tfrac14-\tfrac15\big)\\[2mm] &\,+\cdots\\[2mm] &\,+\big(\tfrac2{n-2}-\tfrac1{n-1}-\tfrac1n\big)\\[2mm] &\,+\big(\tfrac2{n-1}-\tfrac1n-\tfrac1{n+1}\big). \end{align*}

After cancellation,

Sn= 2−12+22 −1n−1n−1n+1= 52−2n−1n+1.\begin{align*} S_n=&\,2-\frac12+\frac22\\[2mm] &\,-\frac1n-\frac1n-\frac1{n+1}\\[4mm] =&\,\frac52-\frac2n-\frac1{n+1}. \end{align*}

Put over the denominator 2n(n+1)2n(n+1):

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

Thus a=5a=5, b=−1b=-1 and c=−4c=-4.

解法二

思路

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把每项拆成两组相邻差。一组的分母相差 11，另一组的分母相差 22，分别抵消后再相加。

答题过程

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From part (a),

 F(r)= (1r−1−1r) +(1r−1−1r+1).\begin{align*} &\,F(r)\\[4mm] =&\,\big(\frac1{r-1}-\frac1r\big)\\[2mm] &\,+\big(\frac1{r-1}-\frac1{r+1}\big). \end{align*}

Let

Un= ∑r=2n(1r−1−1r),Vn= ∑r=2n(1r−1−1r+1).\begin{align*} U_n=&\,\sum_{r=2}^n\big(\frac1{r-1}-\frac1r\big),\\[4mm] V_n=&\,\sum_{r=2}^n\big(\frac1{r-1}-\frac1{r+1}\big). \end{align*}

For the first sum,

Un= (1−12) +(12−13) +(13−14)+⋯ +(1n−2−1n−1) +(1n−1−1n)= 1−1n.\begin{align*} U_n=&\,\big(1-\tfrac12\big)\\[2mm] &\,+\big(\tfrac12-\tfrac13\big)\\[2mm] &\,+\big(\tfrac13-\tfrac14\big)+\cdots\\[2mm] &\,+\big(\tfrac1{n-2}-\tfrac1{n-1}\big)\\[2mm] &\,+\big(\tfrac1{n-1}-\tfrac1n\big)\\[4mm] =&\,1-\frac1n. \end{align*}

For the second sum,

Vn= (1−13) +(12−14) +(13−15)+⋯ +(1n−2−1n) +(1n−1−1n+1)= 1+12−1n−1n+1.\begin{align*} V_n=&\,\big(1-\tfrac13\big)\\[2mm] &\,+\big(\tfrac12-\tfrac14\big)\\[2mm] &\,+\big(\tfrac13-\tfrac15\big)+\cdots\\[2mm] &\,+\big(\tfrac1{n-2}-\tfrac1n\big)\\[2mm] &\,+\big(\tfrac1{n-1}-\tfrac1{n+1}\big)\\[4mm] =&\,1+\frac12-\frac1n-\frac1{n+1}. \end{align*}

Therefore

Sn= Un+Vn= 52−2n−1n+1= 5n(n+1)2n(n+1) −4(n+1)+2n2n(n+1)= 5n2−n−42n(n+1).\begin{align*} S_n=&\,U_n+V_n\\[2mm] =&\,\frac52-\frac2n-\frac1{n+1}\\[4mm] =&\,\frac{5n(n+1)}{2n(n+1)}\\[2mm] &\,-\frac{4(n+1)+2n}{2n(n+1)}\\[4mm] =&\,\frac{5n^2-n-4}{2n(n+1)}. \end{align*}

Thus a=5a=5, b=−1b=-1 and c=−4c=-4.

(c)

解法一

思路

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从 1515 到 2020 的和等于 S20−S14S_{20}-S_{14}。这里 SnS_n 的求和下限是 r=2r=2，所以相减后正好留下所求的六项。

答题过程

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Using part (b),

S20= 5(20)2−20−42(20)(21)= 1976840= 247105,\begin{align*} S_{20}=&\,\frac{5(20)^2-20-4}{2(20)(21)}\\[4mm] =&\,\frac{1976}{840}\\[4mm] =&\,\frac{247}{105}, \end{align*}

and

S14= 5(14)2−14−42(14)(15)= 962420= 481210.\begin{align*} S_{14}=&\,\frac{5(14)^2-14-4}{2(14)(15)}\\[4mm] =&\,\frac{962}{420}\\[4mm] =&\,\frac{481}{210}. \end{align*}

Therefore the required sum is

S20−S14= 247105−481210= 13210.\begin{align*} S_{20}-S_{14}=&\,\frac{247}{105}-\frac{481}{210}\\[4mm] =&\,\frac{13}{210}. \end{align*}