题目
Problem
(a) Use standard results from the list of formulae (MF19) to show that
r=1∑n(2−3r)(5−3r)=an3+bn2+cn,
where a, b and c are integers to be determined.
(3)
(b) Use the method of differences to find r=1∑n(2−3r)(5−3r)1 in terms of n.
(4)
(c) Deduce the value of r=1∑∞(2−3r)(5−3r)1.
(1)
题目中文翻译
(a) 利用公式表（MF19）中的标准结果，证明
r=1∑n(2−3r)(5−3r)=an3+bn2+cn,
其中 a、b 和 c 为待求的整数。
(3)
(b) 用差分法求 r=1∑n(2−3r)(5−3r)1，以 n 表示。
(4)
(c) 由此求 r=1∑∞(2−3r)(5−3r)1 的值。
(1)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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先展开二次乘积，再分别套用平方和、自然数和公式。
答题过程
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Since (2−3r)(5−3r)=9r2−21r+10,
r=1∑n(2−3r)(5−3r)===23n(n+1)(2n+1)−221n(n+1)+10n3n3+29n2+23n−221n2−221n+10n3n3−6n2+n.
Thus a=3, b=−6, c=1.
(b)
解法一
思路
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分式分解后的两个分母相差一个指标，按首尾各三项展开后相邻相消。注意第一项本身为负数。
答题过程
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Partial fractions give
(2−3r)(5−3r)1=31(2−3r1−5−3r1).
For n≥6, the first and last three terms display the cancellation below. For 1≤n≤5, directly cancelling the corresponding terms gives the same result. Therefore, by the method of differences,
r=1∑n(2−3r)(5−3r)1==31[(−11−21)+(−41−−11)+(−71−−41)+⋯+(8−3n1−11−3n1)+(5−3n1−8−3n1)+(2−3n1−5−3n1)]−61+3(2−3n)1.
The interior fractions cancel in adjacent terms.
(c)
解法一
思路
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用(b)的有限和取极限，剩余含n项趋于零。
答题过程
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Taking n→∞ in (b),
r=1∑∞(2−3r)(5−3r)1=−61.
