CalcGospel 國際數學圖譜
返回

IAL 2025 Jan Q4

A Level / Edexcel / FP2

IAL 2025 Jan Paper · Question 4

题目

Problem

(a) Express

r+4r(r+1)(r+2)\begin{align*} \frac{r+4}{r(r+1)(r+2)} \end{align*}

in partial fractions.

(4)

(b) Hence, using the method of differences, show that

∑r=1nr+4r(r+1)(r+2)=n(Pn+Q)2(n+R)(n+S)\begin{align*} \sum_{r=1}^{n} \frac{r+4}{r(r+1)(r+2)} = \frac{n(Pn+Q)}{2(n+R)(n+S)} \end{align*}

where PP, QQ, RR and SS are integers to be found.

(5)

解答

(a)

解法一

思路

展开

分母是三个不同的一次因式，所以设成 Ar+Br+1+Cr+2\frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2}。用代入特殊值可以最快求出系数。

答题过程

展开

Let

r+4r(r+1)(r+2)≡Ar+Br+1+Cr+2\begin{align*} \frac{r+4}{r(r+1)(r+2)} \equiv{}& \frac{A}{r}+\frac{B}{r+1}+\frac{C}{r+2} \end{align*}

Multiply by r(r+1)(r+2)r(r+1)(r+2):

r+4≡A(r+1)(r+2)+Br(r+2)+Cr(r+1)\begin{align*} r+4 \equiv{}& A(r+1)(r+2)+Br(r+2)+Cr(r+1) \end{align*}

Put r=0r=0:

4=2A⇒A=2\begin{align*} 4=2A \quad \Rightarrow \quad A=2 \end{align*}

Put r=−1r=-1:

3=−B⇒B=−3\begin{align*} 3=-B \quad \Rightarrow \quad B=-3 \end{align*}

Put r=−2r=-2:

2=2C⇒C=1\begin{align*} 2=2C \quad \Rightarrow \quad C=1 \end{align*}

Therefore

r+4r(r+1)(r+2)=2r−3r+1+1r+2\begin{align*} \boxed{ \frac{r+4}{r(r+1)(r+2)} = \frac2r-\frac3{r+1}+\frac1{r+2} } \end{align*}

(b)

解法一

思路

展开

把 (a) 的部分分式逐项展开。中间项会大量抵消，最后只留下开头的两项和结尾的两项。这个展开一定要写足够多项，学生才看得出为什么会抵消。

答题过程

展开

Using part (a),

∑r=1nr+4r(r+1)(r+2)=∑r=1n(2r−3r+1+1r+2)\begin{align*} \sum_{r=1}^{n} \frac{r+4}{r(r+1)(r+2)} ={}& \sum_{r=1}^{n} \left(\frac2r-\frac3{r+1}+\frac1{r+2}\right) \end{align*}

Expand the first few and last few terms:

Sn=(2−32+13) +(1−1+14) +(23−34+15) +⋯ +(2n−1−3n+1n+1) +(2n−3n+1+1n+2)\begin{align*} S_n ={}& \left(2-\frac32+\frac13\right)\\[3mm] &\,\hspace{2pt}+\left(1-1+\frac14\right)\\[3mm] &\,\hspace{4pt}+\left(\frac23-\frac34+\frac15\right)\\[3mm] &\,\hspace{6pt}+\cdots\\[3mm] &\,\hspace{8pt}+\left(\frac2{n-1}-\frac3n+\frac1{n+1}\right)\\[3mm] &\,\hspace{10pt}+\left(\frac2n-\frac3{n+1}+\frac1{n+2}\right) \end{align*}

After cancellation,

Sn=2−12−2n+1+1n+2=32−2n+1+1n+2\begin{align*} S_n ={}& 2-\frac12-\frac2{n+1}+\frac1{n+2}\\[3mm] ={}& \frac32-\frac2{n+1}+\frac1{n+2} \end{align*}

Put over a common denominator:

Sn=3(n+1)(n+2)−4(n+2)+2(n+1)2(n+1)(n+2)=3(n2+3n+2)−4n−8+2n+22(n+1)(n+2)=3n2+7n2(n+1)(n+2)=n(3n+7)2(n+1)(n+2)\begin{align*} S_n ={}& \frac{ 3(n+1)(n+2)-4(n+2)+2(n+1) } {2(n+1)(n+2)}\\[3mm] ={}& \frac{ 3(n^2+3n+2)-4n-8+2n+2 } {2(n+1)(n+2)}\\[3mm] ={}& \frac{3n^2+7n}{2(n+1)(n+2)}\\[3mm] ={}& \frac{n(3n+7)}{2(n+1)(n+2)} \end{align*}

Hence

P=3,Q=7,R=1,S=2\begin{align*} \boxed{ P=3,\quad Q=7,\quad R=1,\quad S=2 } \end{align*}

解法二

思路

展开

也可以先把部分分式拆成两组相邻差：

2r−3r+1+1r+2=(2r−2r+1)+(−1r+1+1r+2)\begin{align*} \frac2r-\frac3{r+1}+\frac1{r+2} = \left(\frac2r-\frac2{r+1}\right) +\left(-\frac1{r+1}+\frac1{r+2}\right) \end{align*}

这样每一组都是标准望远镜求和。

答题过程

展开

From part (a),

2r−3r+1+1r+2=(2r−2r+1)+(−1r+1+1r+2)\begin{align*} \frac2r-\frac3{r+1}+\frac1{r+2} ={}& \left(\frac2r-\frac2{r+1}\right) +\left(-\frac1{r+1}+\frac1{r+2}\right) \end{align*}

So

Sn=∑r=1n(2r−2r+1)+∑r=1n(−1r+1+1r+2)\begin{align*} S_n ={}& \sum_{r=1}^{n} \left(\frac2r-\frac2{r+1}\right) +\sum_{r=1}^{n} \left(-\frac1{r+1}+\frac1{r+2}\right) \end{align*}

The two sums telescope:

∑r=1n(2r−2r+1)=2−2n+1\begin{align*} \sum_{r=1}^{n} \left(\frac2r-\frac2{r+1}\right) ={}& 2-\frac2{n+1} \end{align*}

and

∑r=1n(−1r+1+1r+2)=−12+1n+2\begin{align*} \sum_{r=1}^{n} \left(-\frac1{r+1}+\frac1{r+2}\right) ={}& -\frac12+\frac1{n+2} \end{align*}

Hence

Sn=2−2n+1−12+1n+2=32−2n+1+1n+2=n(3n+7)2(n+1)(n+2)\begin{align*} S_n ={}& 2-\frac2{n+1}-\frac12+\frac1{n+2}\\[3mm] ={}& \frac32-\frac2{n+1}+\frac1{n+2}\\[3mm] ={}& \frac{n(3n+7)}{2(n+1)(n+2)} \end{align*}

Therefore

P=3,Q=7,R=1,S=2\begin{align*} \boxed{ P=3,\quad Q=7,\quad R=1,\quad S=2 } \end{align*}