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FP1 2022 June 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)\begin{align*} &\,\sum_{r=0}^{n}(r+1)(r+2)\\ =&\,\frac13(n+1)(n+2)\\ &\,\hspace{2pt}\times(n+3) \end{align*}
(5)

(b) Hence determine the value of

 10×11+11×12 +12×13+… +100×101\begin{align*} &\,10\times11+11\times12\\ &\,\hspace{2pt}+12\times13+\ldots\\ &\,\hspace{4pt}+100\times101 \end{align*}
(3)
题目中文翻译
  1. (a) 利用 ∑r=1nr2\displaystyle\sum_{r=1}^{n} r^2 和 ∑r=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)\begin{align*} &\,\sum_{r=0}^{n}(r+1)(r+2)\\ =&\,\frac13(n+1)(n+2)\\ &\,\hspace{2pt}\times(n+3) \end{align*}
(5)

(b) 由此求

 10×11+11×12 +12×13+… +100×101\begin{align*} &\,10\times11+11\times12\\ &\,\hspace{2pt}+12\times13+\ldots\\ &\,\hspace{4pt}+100\times101 \end{align*}

的值。

(3)

解答

(a)

解法一

思路

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 (r+1)(r+2)= r2+3r+2,\begin{align*} &\,(r+1)(r+2)\\ =&\,r^2+3r+2, \end{align*}

分别套用平方和与等差数列求和公式。注意从 r=0r=0 到 r=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

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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+16[n(2n+1) +9n+12]= n+16 ×(2n2+10n+12)= n+13(n2+5n+6)= n+13[(n2+2n) +(3n+6)]= n+13[n(n+2) +3(n+2)]= 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}{6}\big[ n(2n+1)\\ &\,\hspace{2pt}+9n+12\big]\\ =&\,\frac{n+1}{6}\\ &\,\hspace{2pt}\times(2n^2+10n+12)\\ =&\,\frac{n+1}{3}(n^2+5n+6)\\ =&\,\frac{n+1}{3}\big[(n^2+2n)\\ &\,\hspace{2pt}+(3n+6)\big]\\ =&\,\frac{n+1}{3}\big[n(n+2)\\ &\,\hspace{2pt}+3(n+2)\big]\\ =&\,\frac13(n+1)(n+2)\\ &\,\hspace{2pt}\times(n+3). \end{align*}

This proves the required result.

解法二

思路

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先换指标：令 s=r+1s=r+1，原和式变为从 11 到 n+1n+1 的 s(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

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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)6 ×(2n+3) +(n+1)(n+2)2= (n+1)(n+2) ×(2n+36+12)= (n+1)(n+2) ×2n+3+36= (n+1)(n+2) ×2(n+3)6= 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)}6\\ &\,\hspace{2pt}\times(2n+3)\\ &\,+\frac{(n+1)(n+2)}2\\ =&\,(n+1)(n+2)\\ &\,\hspace{2pt}\times \bigg(\frac{2n+3}{6}+\frac12\bigg)\\ =&\,(n+1)(n+2)\\ &\,\hspace{2pt}\times\frac{2n+3+3}{6}\\ =&\,(n+1)(n+2)\\ &\,\hspace{2pt}\times\frac{2(n+3)}6\\ =&\,\frac13(n+1)(n+2)\\ &\,\hspace{2pt}\times(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) 从 r=10r=10 加到 100100。用从 11 到 100100 的总和减去从 11 到 99 的总和，分别对应 (a) 中的上限 n=99n=99 与 n=8n=8。

答题过程

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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)= 343400−330= 343070.\begin{align*} &\,10\times11+11\times12\\ &\,\hspace{2pt}+12\times13+\cdots\\ &\,\hspace{4pt}+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*}