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IAL 2021 Oct FP1 Q5

A Level / Edexcel / FP1

IAL 2021 Oct Paper · Question 5

题目

Problem

5. (a) Use the standard results for ∑r=1nr3\displaystyle\sum_{r=1}^{n} r^3, ∑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=1nr(r−1)(r−3)=112n(n+1)(n−1)(3n−10)\sum_{r=1}^{n} r(r-1)(r-3) = \frac{1}{12}n(n+1)(n-1)(3n-10)

(5)

(b) Hence show that

∑r=n+12n+1r(r−1)(r−3)=112n(n+1)(an2+bn+c)\sum_{r=n+1}^{2n+1} r(r-1)(r-3) = \frac{1}{12}n(n+1)(an^2 + bn + c)

where aa, bb and cc are integers to be determined.

(3)
题目中文翻译
  1. (a) 利用 ∑r=1nr3\displaystyle\sum_{r=1}^{n} r^3、∑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=1nr(r−1)(r−3)=112n(n+1)(n−1)(3n−10)\sum_{r=1}^{n} r(r-1)(r-3) = \frac{1}{12}n(n+1)(n-1)(3n-10)

(b) 由此证明

∑r=n+12n+1r(r−1)(r−3)=112n(n+1)(an2+bn+c)\sum_{r=n+1}^{2n+1} r(r-1)(r-3) = \frac{1}{12}n(n+1)(an^2 + bn + c)

其中 aa、bb 和 cc 是待定整数。

解答

(a)

解法一

思路

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先展开被求和的代数式，再分别套用三个标准求和公式。提取公因式 112n(n+1)\frac{1}{12}n(n+1) 后，将剩余的二次式因式分解，即可得到题目指定的形式。

答题过程

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First,

r(r−1)(r−3)=r3−4r2+3r.r(r-1)(r-3)=r^3-4r^2+3r.

Using the standard results,

 ∑r=1nr(r−1)(r−3)= ∑r=1nr3− 4∑r=1nr2+ 3∑r=1nr= n2(n+1)24− 2n(n+1)(2n+1)3+ 3n(n+1)2= 112n(n+1)[3n(n+1)−8(2n+1)+18]= 112n(n+1)(3n2−13n+10)= 112n(n+1)(n−1)(3n−10).\begin{align*} &\,\sum_{r=1}^{n}r(r-1)(r-3) \\ =&\,\sum_{r=1}^{n}r^3 \\ -&\,\hspace{2pt}4\sum_{r=1}^{n}r^2 \\ +&\,\hspace{4pt}3\sum_{r=1}^{n}r \\ =&\,\frac{n^2(n+1)^2}{4} \\ -&\,\hspace{2pt}\frac{2n(n+1)(2n+1)}{3} \\ +&\,\hspace{4pt}\frac{3n(n+1)}{2} \\ =&\,\frac{1}{12}n(n+1) \big[3n(n+1)-8(2n+1)+18\big] \\ =&\,\frac{1}{12}n(n+1) (3n^2-13n+10) \\ =&\,\boxed{\frac{1}{12}n(n+1)(n-1)(3n-10)}. \end{align*}

(b)

解法一

思路

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承接 (a)，把从 11 到 mm 的和记为 SmS_m。所求区间和等于 S2n+1−SnS_{2n+1}-S_n；代入 (a) 的结果并化简，即可比较系数求出 aa、bb、cc。

答题过程

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Let

Sm=∑r=1mr(r−1)(r−3).S_m=\sum_{r=1}^{m}r(r-1)(r-3).

From part (a),

Sm=112m(m+1)(m−1)(3m−10).S_m=\frac{1}{12}m(m+1)(m-1)(3m-10).

Therefore,

 ∑r=n+12n+1r(r−1)(r−3)= S2n+1−Sn= 112(2n+1)(2n+2) ⋅(2n)(6n−7)− 112n(n+1)(n−1)(3n−10)= 112n(n+1)[4(2n+1)(6n−7) −(n−1)(3n−10)]= 112n(n+1)(45n2−19n−38).\begin{align*} &\,\sum_{r=n+1}^{2n+1}r(r-1)(r-3) \\ =&\,S_{2n+1}-S_n \\ =&\,\frac{1}{12}(2n+1)(2n+2) \\ &\,\hspace{2pt}\cdot(2n)(6n-7) \\ -&\,\hspace{4pt}\frac{1}{12}n(n+1)(n-1)(3n-10) \\ =&\,\frac{1}{12}n(n+1) \big[4(2n+1)(6n-7) \\ &\,\hspace{6pt}-(n-1)(3n-10)\big] \\ =&\,\frac{1}{12}n(n+1) (45n^2-19n-38). \end{align*}

Hence

a=45,b=−19,c=−38.\boxed{a=45,\qquad b=-19,\qquad c=-38}.