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FP2 2021 June Q1

A Level / Edexcel / FP2

IAL 2021 June Paper · Question 1

题目

Problem

1. (a) Express

 2r(r2−1)\begin{align*} &\,\frac{2}{r(r^2 - 1)} \end{align*}

in partial fractions.

(3)

(b) Hence find, in terms of nn,

 ∑r=2n1r(r2−1)\begin{align*} &\,\sum_{r=2}^{n} \frac{1}{r(r^2 - 1)} \end{align*}

Give your answer in the form

 n2+An+BCn(n+1)\begin{align*} &\,\frac{n^2 + An + B}{Cn(n + 1)} \end{align*}

where AA, BB and CC are constants to be found.

(5)
题目中文翻译

1. (a) 将

 2r(r2−1)\begin{align*} &\,\frac{2}{r(r^2-1)} \end{align*}

分解为部分分式。

(3)

(b) 由此用 nn 表示

 ∑r=2n1r(r2−1)\begin{align*} &\,\sum_{r=2}^{n}\frac{1}{r(r^2-1)} \end{align*}

将答案写成

 n2+An+BCn(n+1)\begin{align*} &\,\frac{n^2+An+B}{Cn(n+1)} \end{align*}

的形式，其中 AA、BB 和 CC 是待求常数。

(5)

解答

(a)

解法一

思路

展开

先把分母因式分解：

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

然后设成三个简单分式，再通过通分后的恒等式求系数。这里的分式结构会在 (b) 产生相邻项抵消。

答题过程

展开

Let

 2r(r2−1)= Ar−1+Br +Cr+1.\begin{align*} &\,\frac2{r(r^2-1)}\\[4mm] =&\,\frac A{r-1}+\frac Br\\[4mm] &\,\hspace{2pt}+\frac C{r+1}. \end{align*}

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

2= Ar(r+1) +B(r−1)(r+1) +Cr(r−1).\begin{align*} 2=&\,Ar(r+1)\\[4mm] &\,\hspace{2pt}+B(r-1)(r+1)\\[4mm] &\,\hspace{4pt}+Cr(r-1). \end{align*}

Substitute convenient values of rr into this identity:

r=1:2= 2A,A= 1,r=0:2= −B,B= −2,r=−1:2= 2C,C= 1.\begin{align*} r=1:\quad 2=&\,2A,\\[4mm] A=&\,1,\\[4mm] r=0:\quad 2=&\,-B,\\[4mm] B=&\,-2,\\[4mm] r=-1:\quad 2=&\,2C,\\[4mm] C=&\,1. \end{align*}

Therefore,

 2r(r2−1)= 1r−1−2r +1r+1.\begin{align*} &\,\frac2{r(r^2-1)}\\[4mm] =&\,\frac1{r-1}-\frac2r\\[4mm] &\,\hspace{2pt}+\frac1{r+1}. \end{align*}

(b)

解法一

思路

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(a) 分解的是题目所求通项的两倍，所以求和后还要乘 12\dfrac12。把开头三项和最后两项完整写出，显示中间项如何抵消，再通分成题目要求的形式。

答题过程

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From part (a), define

Tr= 1r−1−2r +1r+1.\begin{align*} T_r=&\,\frac1{r-1}-\frac2r\\[4mm] &\,\hspace{2pt}+\frac1{r+1}. \end{align*}

Then

1r(r2−1)=12Tr.\frac1{r(r^2-1)}=\frac12T_r.

Let

Sn=∑r=2n1r(r2−1).S_n=\sum_{r=2}^n\frac1{r(r^2-1)}.

Hence

Sn=12∑r=2nTr.S_n=\frac12\sum_{r=2}^nT_r.

Writing out the first three and last two terms,

 ∑r=2nTr= (1−22+13) +(12−23+14) +(13−24+15) +⋯ +(1n−2−2n−1 +1n) +(1n−1−2n +1n+1).\begin{align*} &\,\sum_{r=2}^nT_r\\[4mm] =&\,\bigg(1-\frac22+\frac13\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac12-\frac23+\frac14\bigg)\\[4mm] &\,\hspace{4pt}+\bigg(\frac13-\frac24+\frac15\bigg)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{2pt}+\bigg(\frac1{n-2}-\frac2{n-1}\\[4mm] &\,\hspace{4pt}+\frac1n\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac1{n-1}-\frac2n\\[4mm] &\,\hspace{4pt}+\frac1{n+1}\bigg). \end{align*}

The middle terms cancel, leaving

 ∑r=2nTr= 1−22+12 +1n−2n+1n+1= 12−1n+1n+1.\begin{align*} &\,\sum_{r=2}^nT_r\\[4mm] =&\,1-\frac22+\frac12\\[4mm] &\,\hspace{2pt}+\frac1n-\frac2n+\frac1{n+1}\\[4mm] =&\,\frac12-\frac1n+\frac1{n+1}. \end{align*}

Therefore,

Sn= 12(12−1n +1n+1)= 14−12n +12(n+1).\begin{align*} S_n=&\,\frac12\bigg(\frac12-\frac1n\\[4mm] &\,\hspace{2pt}+\frac1{n+1}\bigg)\\[4mm] =&\,\frac14-\frac1{2n}\\[4mm] &\,\hspace{2pt}+\frac1{2(n+1)}. \end{align*}

Using the common denominator 4n(n+1)4n(n+1), let the numerator be NN. Then

N= n(n+1)−2(n+1) +2n= n2+n−2n−2+2n= n2+n−2.\begin{align*} N=&\,n(n+1)-2(n+1)\\[4mm] &\,\hspace{2pt}+2n\\[4mm] =&\,n^2+n-2n-2+2n\\[4mm] =&\,n^2+n-2. \end{align*}

Hence

Sn=n2+n−24n(n+1).S_n=\frac{n^2+n-2}{4n(n+1)}.

Comparing with the required form,

A= 1,B= −2,C= 4.\begin{align*} A=&\,1,\\[4mm] B=&\,-2,\\[4mm] C=&\,4. \end{align*}