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IAL 2021 June Q1

A Level / Edexcel / FP2

IAL 2021 June Paper · Question 1

题目

Problem

(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)

解答

(a)

解法一

思路

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先把分母因式分解：

r(r2−1)=r(r−1)(r+1).\begin{align*} r(r^2-1)=r(r-1)(r+1). \end{align*}

然后设成三个简单分式。这里的分式结构会在 (b) 产生相邻项抵消。

答题过程

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Let

2r(r2−1)= Ar−1+Br+Cr+1.\begin{align*} \frac{2}{r(r^2-1)} =&\,\frac{A}{r-1}+\frac{B}{r}+\frac{C}{r+1}. \end{align*}

Since r(r2−1)=r(r−1)(r+1)r(r^2-1)=r(r-1)(r+1),

2= Ar(r+1)+B(r−1)(r+1)+Cr(r−1).\begin{align*} 2 =&\,Ar(r+1)+B(r-1)(r+1)+Cr(r-1). \end{align*}

Substitute convenient values of rr:

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

Therefore,

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

(b)

解法一

思路

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题目问的是

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

而 (a) 分解的是两倍的分式，所以最后要乘 12\dfrac12。把首几项和末几项写出来，中间项会望远镜式抵消。

答题过程

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From part (a),

1r(r2−1)= 12(1r−1−2r+1r+1).\begin{align*} \frac{1}{r(r^2-1)} =&\,\frac12\left( \frac{1}{r-1}-\frac2r+\frac{1}{r+1} \right). \end{align*}

Hence,

 ∑r=2n1r(r2−1)= 12∑r=2n(1r−1−2r+1r+1).\begin{align*} &\,\sum_{r=2}^{n}\frac{1}{r(r^2-1)}\\[4mm] =&\,\frac12\sum_{r=2}^{n} \left( \frac{1}{r-1}-\frac2r+\frac{1}{r+1} \right). \end{align*}

Write out the terms:

 ∑r=2n(1r−1−2r+1r+1)= (1−22+13) +(12−23+14) +(13−24+15) +⋯ +(1n−2−2n−1 +1n) +(1n−1−2n +1n+1).\begin{align*} &\,\sum_{r=2}^{n} \left( \frac{1}{r-1}-\frac2r+\frac{1}{r+1} \right)\\[4mm] =&\,\left(1-\frac22+\frac13\right)\\[4mm] &\,\hspace{2pt}+\left(\frac12-\frac23+\frac14\right)\\[4mm] &\,\hspace{4pt}+\left(\frac13-\frac24+\frac15\right)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{8pt}+\left(\frac{1}{n-2}-\frac{2}{n-1}\right.\\[2mm] &\,\hspace{10pt}\left.+\frac1n\right)\\[4mm] &\,\hspace{12pt}+\left(\frac{1}{n-1}-\frac2n\right.\\[2mm] &\,\hspace{14pt}\left.+\frac{1}{n+1}\right). \end{align*}

The middle terms cancel, leaving

 ∑r=2n(1r−1−2r+1r+1)= 12−1n+1n+1.\begin{align*} &\,\sum_{r=2}^{n} \left( \frac{1}{r-1}-\frac2r+\frac{1}{r+1} \right)\\[4mm] =&\,\frac12-\frac1n+\frac{1}{n+1}. \end{align*}

Therefore,

 ∑r=2n1r(r2−1)= 12(12−1n+1n+1)= 14−12n+12(n+1).\begin{align*} &\,\sum_{r=2}^{n}\frac{1}{r(r^2-1)}\\[4mm] =&\,\frac12\left( \frac12-\frac1n+\frac{1}{n+1} \right)\\[4mm] =&\,\frac14-\frac{1}{2n}+\frac{1}{2(n+1)}. \end{align*}

Putting this over the required denominator,

 14−12n+12(n+1)= n(n+1)−2(n+1)+2n4n(n+1)= n2+n−24n(n+1).\begin{align*} &\,\frac14-\frac{1}{2n}+\frac{1}{2(n+1)}\\[4mm] =&\,\frac{n(n+1)-2(n+1)+2n} {4n(n+1)}\\[4mm] =&\,\frac{n^2+n-2}{4n(n+1)}. \end{align*}

Thus

A=1,B=−2,C=4.\begin{align*} A=1,\qquad B=-2,\qquad C=4. \end{align*}