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FP2 2023 Jan Q2

A Level / Edexcel / FP2

IAL 2023 Jan Paper · Question 2

题目

Problem

2.

(a) Express

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

in partial fractions.

(2)

(b) Hence, using the method of differences, show that for all integer values of nn,

∑r=1n1(2r−1)(2r+1)(2r+3)= n(n+2)a(2n+b)(2n+c)\begin{align*} \sum_{r=1}^{n} \frac{1}{(2r - 1)(2r + 1)(2r + 3)} =&\, \frac{n(n + 2)}{a(2n + b)(2n + c)} \end{align*}

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

(4)
题目中文翻译

(a) 将

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

分解为部分分式。

(2)

(b) 由此利用逐项相消法，证明对所有整数 nn，

∑r=1n1(2r−1)(2r+1)(2r+3)= n(n+2)a(2n+b)(2n+c), \begin{align*} \sum_{r=1}^n\frac{1}{(2r-1)(2r+1)(2r+3)} =&\,\frac{n(n+2)}{a(2n+b)(2n+c)}, \end{align*}

其中 aa、bb、cc 为待求的整数。

(4)

解答

(a)

解法一

思路

展开

分母由三个一次因式组成，所以设成三个简单分式。代入使其中一个因式为零的 nn 值，可以快速求出系数。

答题过程

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Let

1(2n−1)(2n+1)(2n+3)= A2n−1+B2n+1+C2n+3.\begin{align*} \frac{1}{(2n-1)(2n+1)(2n+3)} =&\,\frac{A}{2n-1}+\frac{B}{2n+1} +\frac{C}{2n+3}. \end{align*}

Multiplying by (2n−1)(2n+1)(2n+3)(2n-1)(2n+1)(2n+3),

1= A(2n+1)(2n+3) +B(2n−1)(2n+3) +C(2n−1)(2n+1).\begin{align*} 1=&\,A(2n+1)(2n+3)\\[2mm] &\,\hspace{2pt}+B(2n-1)(2n+3)\\[2mm] &\,\hspace{4pt}+C(2n-1)(2n+1). \end{align*}

Put n=12n=\frac{1}{2}:

1=A(2)(4)⟹A=18.\begin{align*} 1=A(2)(4) \quad\Longrightarrow\quad A=\frac{1}{8}. \end{align*}

Put n=−12n=-\frac{1}{2}:

1=B(−2)(2)⟹B=−14.\begin{align*} 1=B(-2)(2) \quad\Longrightarrow\quad B=-\frac{1}{4}. \end{align*}

Put n=−32n=-\frac{3}{2}:

1=C(−4)(−2)⟹C=18.\begin{align*} 1=C(-4)(-2) \quad\Longrightarrow\quad C=\frac{1}{8}. \end{align*}

Therefore

1(2n−1)(2n+1)(2n+3)=18(2n−1)−14(2n+1)+18(2n+3).\begin{align*} \boxed{ \frac{1}{(2n-1)(2n+1)(2n+3)} =\frac{1}{8(2n-1)} -\frac{1}{4(2n+1)} +\frac{1}{8(2n+3)} }. \end{align*}

(b)

解法一

思路

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把 (a) 的结果写成

18(12r−1−22r+1+12r+3).\begin{align*} \frac{1}{8}\bigg(\frac{1}{2r-1} -\frac{2}{2r+1} +\frac{1}{2r+3}\bigg). \end{align*}

然后展开前几项和最后几项，保留下没有抵消的项。

答题过程

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From part (a),

1(2r−1)(2r+1)(2r+3)= 18(12r−1−22r+1+12r+3).\begin{align*} \frac{1}{(2r-1)(2r+1)(2r+3)} =&\,\frac{1}{8}\bigg(\frac{1}{2r-1} -\frac{2}{2r+1} +\frac{1}{2r+3}\bigg). \end{align*}

So

 ∑r=1n1(2r−1)(2r+1)(2r+3)= 18∑r=1n(12r−1−22r+1+12r+3).\begin{align*} &\,\sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)(2r+3)}\\[2mm] =&\,\frac{1}{8}\sum_{r=1}^{n} \bigg(\frac{1}{2r-1}-\frac{2}{2r+1} +\frac{1}{2r+3}\bigg). \end{align*}

Write out the terms inside the bracket:

 (1−23+15) +(13−25+17) +(15−27+19) +⋯ +(12n−3−22n−1+12n+1) +(12n−1−22n+1+12n+3).\begin{align*} &\,\bigg(1-\frac{2}{3}+\frac{1}{5}\bigg)\\[4mm] &\,\hspace{2pt}+\bigg(\frac{1}{3}-\frac{2}{5}+\frac{1}{7}\bigg)\\[4mm] &\,\hspace{4pt}+\bigg(\frac{1}{5}-\frac{2}{7}+\frac{1}{9}\bigg)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{8pt}+\bigg(\frac{1}{2n-3} -\frac{2}{2n-1}+\frac{1}{2n+1}\bigg)\\[4mm] &\,\hspace{10pt}+\bigg(\frac{1}{2n-1} -\frac{2}{2n+1}+\frac{1}{2n+3}\bigg). \end{align*}

After cancellation,

∑r=1n1(2r−1)(2r+1)(2r+3)= 18(1−13−12n+1+12n+3)= 18(23−2(2n+1)(2n+3)).\begin{align*} \sum_{r=1}^{n} \frac{1}{(2r-1)(2r+1)(2r+3)} =&\,\frac{1}{8} \bigg(1-\frac{1}{3} -\frac{1}{2n+1}+\frac{1}{2n+3}\bigg)\\[2mm] =&\,\frac{1}{8} \bigg(\frac{2}{3} -\frac{2}{(2n+1)(2n+3)}\bigg). \end{align*}

Combine into a single fraction:

18(23−2(2n+1)(2n+3))= 18(2(2n+1)(2n+3)−63(2n+1)(2n+3))= 18(2(4n2+8n+3)−63(2n+1)(2n+3))= 18(8n2+16n3(2n+1)(2n+3))= n(n+2)3(2n+1)(2n+3).\begin{align*} \frac{1}{8} \bigg(\frac{2}{3} -\frac{2}{(2n+1)(2n+3)}\bigg) =&\,\frac{1}{8} \bigg( \frac{2(2n+1)(2n+3)-6} {3(2n+1)(2n+3)} \bigg)\\[2mm] =&\,\frac{1}{8} \bigg( \frac{2(4n^2+8n+3)-6} {3(2n+1)(2n+3)} \bigg)\\[2mm] =&\,\frac{1}{8} \bigg( \frac{8n^2+16n} {3(2n+1)(2n+3)} \bigg)\\[2mm] =&\,\frac{n(n+2)}{3(2n+1)(2n+3)}. \end{align*}

Thus

a=3,b=1,c=3.\begin{align*} \boxed{a=3,\qquad b=1,\qquad c=3}. \end{align*}