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IAL 2022 June FP1 Q8

A Level / Edexcel / FP1

IAL 2022 June Paper · Question 8

题目

Problem

8. (a) Use the standard results for r=1nr2\displaystyle\sum_{r=1}^{n} r^2 and r=1nr\displaystyle\sum_{r=1}^{n} r to show that for all positive integers nn

r=0n(r+1)(r+2)=13(n+1)(n+2)(n+3)\sum_{r=0}^{n} (r+1)(r+2) = \frac{1}{3}(n+1)(n+2)(n+3)

(5)

(b) Hence determine the value of

10×11+11×12+12×13++100×10110 \times 11 + 11 \times 12 + 12 \times 13 + \ldots + 100 \times 101

(3)
题目中文翻译
  1. (a) 利用 r=1nr2\displaystyle\sum_{r=1}^{n} r^2r=1nr\displaystyle\sum_{r=1}^{n} r 的标准结果证明,对于所有正整数 nn

r=0n(r+1)(r+2)=13(n+1)(n+2)(n+3)\sum_{r=0}^{n} (r+1)(r+2) = \frac{1}{3}(n+1)(n+2)(n+3)

(b) 由此求

10×11+11×12+12×13++100×10110 \times 11 + 11 \times 12 + 12 \times 13 + \ldots + 100 \times 101

的值。

解答

(a)

解法一

思路

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先展开 (r+1)(r+2)=r2+3r+2(r+1)(r+2)=r^2+3r+2,分别套用平方和与等差数列求和公式。注意从 r=0r=0r=nr=n 一共有 n+1n+1 个常数项,最后因式分解至题目指定形式。

答题过程

展开

Using the standard results,

r=0n(r+1)(r+2)=r=0n(r2+3r+2)=n(n+1)(2n+1)6+3n(n+1)2+2(n+1)=(n+1)[n(2n+1)6+3n2+2]=n+13(n2+5n+6)=13(n+1)(n+2)(n+3).\begin{align*} &\,\sum_{r=0}^{n}(r+1)(r+2)\\ =&\,\sum_{r=0}^{n}(r^2+3r+2)\\ =&\,\frac{n(n+1)(2n+1)}6\\ &\,\hspace{2pt}+\frac{3n(n+1)}2\\ &\,\hspace{4pt}+2(n+1)\\ =&\,(n+1)\bigg[ \frac{n(2n+1)}6\\ &\,\hspace{2pt}+\frac{3n}{2}+2 \bigg]\\ =&\,\frac{n+1}{3}(n^2+5n+6)\\ =&\,\boxed{ \frac13(n+1)(n+2)(n+3)}. \end{align*}

This proves the required result.

解法二

思路

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官方第二种路线先换指标:令 s=r+1s=r+1,原和式变为从 11n+1n+1s(s+1)s(s+1) 之和。这样没有额外常数项计数,直接使用上限为 n+1n+1 的两个标准公式。

答题过程

展开

Let s=r+1s=r+1. As rr runs from 00 to nn, ss runs from 11 to n+1n+1. Therefore,

r=0n(r+1)(r+2)=s=1n+1s(s+1)=s=1n+1(s2+s)=(n+1)(n+2)(2n+3)6+(n+1)(n+2)2=(n+1)(n+2)(2n+36+12)=13(n+1)(n+2)(n+3).\begin{align*} &\,\sum_{r=0}^{n}(r+1)(r+2)\\ =&\,\sum_{s=1}^{n+1}s(s+1)\\ =&\,\sum_{s=1}^{n+1}(s^2+s)\\ =&\,\frac{(n+1)(n+2)(2n+3)}6\\ &\,+\frac{(n+1)(n+2)}2\\ =&\,(n+1)(n+2) \bigg(\frac{2n+3}{6}\\ &\,\hspace{2pt}+\frac12\bigg)\\ =&\,\boxed{ \frac13(n+1)(n+2)(n+3)}. \end{align*}

(b)

解法一

思路

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承接 (a),目标是 r(r+1)r(r+1)r=10r=10 加到 100100。用从 11100100 的总和减去从 1199 的总和,分别对应 (a) 中的上限 n=99n=99n=8n=8

答题过程

展开

Using the result from part (a),

10×11+11×12+12×13++100×101=r=099(r+1)(r+2)r=08(r+1)(r+2)=13(100)(101)(102)13(9)(10)(11)=343400330=343070.\begin{align*} &\,10\times11+11\times12\\ &\,\hspace{2pt}+12\times13+\cdots+100\times101\\ =&\,\sum_{r=0}^{99}(r+1)(r+2)\\ &\,\hspace{2pt}-\sum_{r=0}^{8}(r+1)(r+2)\\ =&\,\frac13(100)(101)(102)\\ &\,\hspace{2pt}-\frac13(9)(10)(11)\\ =&\,343400-330\\ =&\,\boxed{343070}. \end{align*}