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IAL 2024 June Q3

A Level / Edexcel / FP2

IAL 2024 June Paper · Question 3

题目

Problem

(a) Express

1(n+3)(n+5)\begin{align*} \frac{1}{(n + 3)(n + 5)} \end{align*}

in partial fractions.

(2)

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

∑r=1n1(r+3)(r+5)= n(pn+q)40(n+4)(n+5)\begin{align*} \sum_{r=1}^{n} \frac{1}{(r + 3)(r + 5)} =&\, \frac{n(pn + q)}{40(n + 4)(n + 5)} \end{align*}

where pp and qq are integers to be determined.

(4)

(c) Use the answer to part (b) to determine, as a simplified fraction, the value of

19×11+110×12+⋯+124×26\begin{align*} \frac{1}{9 \times 11} + \frac{1}{10 \times 12} + \dots + \frac{1}{24 \times 26} \end{align*}
(2)

解答

(a)

解法一

思路

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分母两个一次因式相差 22，部分分式会是两个简单分式相减。

答题过程

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Let

1(n+3)(n+5)≡An+3+Bn+5\begin{align*} \frac{1}{(n+3)(n+5)} \equiv{}& \frac{A}{n+3}+\frac{B}{n+5} \end{align*}

Then

1≡A(n+5)+B(n+3)\begin{align*} 1\equiv A(n+5)+B(n+3) \end{align*}

Putting n=−3n=-3 gives 1=2A1=2A, so A=12A=\frac12.

Putting n=−5n=-5 gives 1=−2B1=-2B, so B=−12B=-\frac12.

Therefore

1(n+3)(n+5)=12(n+3)−12(n+5)\begin{align*} \boxed{ \frac{1}{(n+3)(n+5)} =\frac{1}{2(n+3)}-\frac{1}{2(n+5)} } \end{align*}

(b)

解法一

思路

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把 (a) 的结果用于求和后，展开前几项和最后几项。由于分母相差 22，会留下前两个正项和最后两个负项。

答题过程

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Using part (a),

∑r=1n1(r+3)(r+5)= 12∑r=1n(1r+3−1r+5)\begin{align*} \sum_{r=1}^{n}\frac{1}{(r+3)(r+5)} =&\, \frac12\sum_{r=1}^{n} \left(\frac{1}{r+3}-\frac{1}{r+5}\right) \end{align*}

Expanding the terms,

12∑r=1n(1r+3−1r+5)= 12(14−16) +12(15−17) +12(16−18) +⋯ +12(1n+2−1n+4) +12(1n+3−1n+5)\begin{align*} \frac12\sum_{r=1}^{n} \left(\frac{1}{r+3}-\frac{1}{r+5}\right) =&\, \frac12\left(\frac14-\frac16\right)\\[4mm] &\,\hspace{2pt}+\frac12\left(\frac15-\frac17\right)\\[4mm] &\,\hspace{4pt}+\frac12\left(\frac16-\frac18\right)\\[4mm] &\,\hspace{6pt}+\cdots\\[4mm] &\,\hspace{8pt}+\frac12\left(\frac{1}{n+2}-\frac{1}{n+4}\right)\\[4mm] &\,\hspace{10pt}+\frac12\left(\frac{1}{n+3}-\frac{1}{n+5}\right) \end{align*}

After cancellation,

∑r=1n1(r+3)(r+5)= 12(14+15−1n+4−1n+5)= 12(920−2n+9(n+4)(n+5))= 12(9(n+4)(n+5)−20(2n+9)20(n+4)(n+5))= 9(n2+9n+20)−40n−18040(n+4)(n+5)= 9n2+41n40(n+4)(n+5)= n(9n+41)40(n+4)(n+5)\begin{align*} \sum_{r=1}^{n}\frac{1}{(r+3)(r+5)} =&\, \frac12\left( \frac14+\frac15-\frac{1}{n+4}-\frac{1}{n+5} \right)\\[4mm] =&\, \frac12\left( \frac{9}{20}-\frac{2n+9}{(n+4)(n+5)} \right)\\[4mm] =&\, \frac12\left( \frac{9(n+4)(n+5)-20(2n+9)} {20(n+4)(n+5)} \right)\\[4mm] =&\, \frac{9(n^2+9n+20)-40n-180} {40(n+4)(n+5)}\\[4mm] =&\, \frac{9n^2+41n}{40(n+4)(n+5)}\\[4mm] =&\, \frac{n(9n+41)}{40(n+4)(n+5)} \end{align*}

Hence

p=9,q=41\begin{align*} \boxed{p=9,\qquad q=41} \end{align*}

(c)

解法一

思路

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9×119\times11 对应 (r+3)(r+5)(r+3)(r+5) 中的 r=6r=6，而 24×2624\times26 对应 r=21r=21。所以用 S21−S5S_{21}-S_5。

答题过程

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Let

Sn=∑r=1n1(r+3)(r+5)=n(9n+41)40(n+4)(n+5)\begin{align*} S_n=\sum_{r=1}^{n}\frac{1}{(r+3)(r+5)} =\frac{n(9n+41)}{40(n+4)(n+5)} \end{align*}

Then the required sum is

S21−S5= 21(9⋅21+41)40(25)(26)−5(9⋅5+41)40(9)(10)= 21(230)26000−5(86)3600= 4832600−43360= 4347−279523400= 155223400= 1942925\begin{align*} S_{21}-S_5 =&\, \frac{21(9\cdot21+41)}{40(25)(26)} -\frac{5(9\cdot5+41)}{40(9)(10)}\\[4mm] =&\, \frac{21(230)}{26000} -\frac{5(86)}{3600}\\[4mm] =&\, \frac{483}{2600}-\frac{43}{360}\\[4mm] =&\, \frac{4347-2795}{23400}\\[4mm] =&\, \frac{1552}{23400}\\[4mm] =&\, \boxed{\frac{194}{2925}} \end{align*}