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FP1 2021 June Q7

A Level / Edexcel / FP1

IAL 2021 June Paper · Question 7

题目

Problem

7. (a) Prove by induction that for n∈Nn \in \mathbb{N}

 ∑r=1nr2= 16n(n+1) ×(2n+1)\begin{align*} &\,\sum_{r=1}^{n}r^2\\ =&\,\frac16n(n+1)\\ &\,\hspace{2pt}\times(2n+1) \end{align*}
(5)

(b) Hence show that

 ∑r=1n(r2+2)= n6(an2+bn+c)\begin{align*} &\,\sum_{r=1}^{n}(r^2+2)\\ =&\,\frac n6(an^2+bn+c) \end{align*}

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

(4)

(c) Using your answers to part (b), find the value of

∑r=1025(r2+2)\sum_{r=10}^{25} (r^2+2)
(2)
题目中文翻译
  1. (a) 用数学归纳法证明，对于 n∈Nn \in \mathbb{N}
 ∑r=1nr2= 16n(n+1) ×(2n+1)\begin{align*} &\,\sum_{r=1}^{n}r^2\\ =&\,\frac16n(n+1)\\ &\,\hspace{2pt}\times(2n+1) \end{align*}
(5)

(b) 由此证明

 ∑r=1n(r2+2)= n6(an2+bn+c)\begin{align*} &\,\sum_{r=1}^{n}(r^2+2)\\ =&\,\frac n6(an^2+bn+c) \end{align*}

其中 aa、bb 和 cc 是待求的整数。

(4)

(c) 利用 (b) 的答案，求

∑r=1025(r2+2)\sum_{r=10}^{25} (r^2+2)

的值。

(2)

解答

(a)

解法一

思路

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先验证 n=1n=1。再假设公式对 n=kn=k 成立，在前 kk 项平方和上加上第 k+1k+1 项的平方；整理并因式分解成把 nn 替换为 k+1k+1 后的目标形式，最后写出完整归纳结论。

答题过程

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For n=1n=1,

LHS=12=1,\text{LHS}=1^2=1,

and

RHS=16(1)(2)(3)=1.\text{RHS}=\frac16(1)(2)(3)=1.

Thus the result is true for n=1n=1.

Assume that the result is true for n=kn=k, where k∈Nk\in\mathbb{N}. Then

 ∑r=1kr2= 16k(k+1)(2k+1).\begin{align*} &\,\sum_{r=1}^{k}r^2\\ =&\,\frac16k(k+1)(2k+1). \end{align*}

For n=k+1n=k+1,

 ∑r=1k+1r2= ∑r=1kr2 +(k+1)2= 16k(k+1)(2k+1) +(k+1)2= k+16 ×[k(2k+1) +6(k+1)]= k+16 ×(2k2+k+6k+6)= k+16 ×(2k2+7k+6)= 16(k+1) ×(k+2)(2k+3)= 16(k+1) ×((k+1)+1) ×(2(k+1)+1).\begin{align*} &\,\sum_{r=1}^{k+1}r^2\\ =&\,\sum_{r=1}^{k}r^2\\ &\,\hspace{2pt}+(k+1)^2\\ =&\,\frac16k(k+1)(2k+1)\\ &\,\hspace{2pt}+(k+1)^2\\ =&\,\frac{k+1}{6}\\ &\,\hspace{2pt}\times\big[k(2k+1)\\ &\,\hspace{4pt}+6(k+1)\big]\\ =&\,\frac{k+1}{6}\\ &\,\hspace{2pt}\times\big(2k^2+k+6k+6\big)\\ =&\,\frac{k+1}{6}\\ &\,\hspace{2pt}\times\big(2k^2+7k+6\big)\\ =&\,\frac16(k+1)\\ &\,\hspace{2pt}\times(k+2)(2k+3)\\ =&\,\frac16(k+1)\\ &\,\hspace{2pt}\times\big((k+1)+1\big)\\ &\,\hspace{4pt}\times\big(2(k+1)+1\big). \end{align*}

Here the quadratic factorises as

 2k2+7k+6= 2k2+4k+3k+6= 2k(k+2)+3(k+2)= (k+2)(2k+3).\begin{align*} &\,2k^2+7k+6\\ =&\,2k^2+4k+3k+6\\ =&\,2k(k+2)+3(k+2)\\ =&\,(k+2)(2k+3). \end{align*}

Therefore, if the result is true for n=kn=k, it is true for n=k+1n=k+1. Since it is true for n=1n=1, the result is true for all n∈Nn\in\mathbb{N} by mathematical induction.

(b)

解法一

思路

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承接 (a)，把和式拆成平方和与 nn 个常数 22 的和。代入 (a) 的公式并提取 n/6n/6，再与题目给出的系数形式比较。

答题过程

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Using the result from part (a),

 ∑r=1n(r2+2)= ∑r=1nr2 +2n= 16n(n+1)(2n+1) +2n= n6 ×[(n+1)(2n+1) +12]= n6 ×(2n2+n+2n +1+12)= n6(2n2+3n+13).\begin{align*} &\,\sum_{r=1}^{n}(r^2+2)\\ =&\,\sum_{r=1}^{n}r^2\\ &\,\hspace{2pt}+2n\\ =&\,\frac16n(n+1)(2n+1)\\ &\,\hspace{2pt}+2n\\ =&\,\frac n6\\ &\,\hspace{2pt}\times\big[(n+1)(2n+1)\\ &\,\hspace{4pt}+12\big]\\ =&\,\frac n6\\ &\,\hspace{2pt}\times\big(2n^2+n+2n\\ &\,\hspace{4pt}+1+12\big)\\ =&\,\frac n6\big(2n^2+3n+13\big). \end{align*}

Comparing this with

n6(an2+bn+c),\frac n6(an^2+bn+c),

we obtain

a= 2,b= 3,c= 13.\begin{align*} a=&\,\boxed{2},\\ b=&\,\boxed{3},\\ c=&\,\boxed{13}. \end{align*}

(c)

解法一

思路

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使用 (b) 的结果，把从 r=10r=10 到 2525 的和写成前 2525 项之和减去前 99 项之和。必须减到 99，这样才会保留从 1010 开始的所有项。

答题过程

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Let

Sn= ∑r=1n(r2+2)= n6(2n2+3n+13).\begin{align*} S_n=&\,\sum_{r=1}^{n}(r^2+2)\\ =&\,\frac n6\big(2n^2+3n+13\big). \end{align*}

Using part (b),

 ∑r=1025(r2+2)= S25−S9= 256 ×(2(25)2+3(25) +13) −96 ×(2(9)2+3(9) +13)= 256(1338) −96(202)= 5575−303= 5272.\begin{align*} &\,\sum_{r=10}^{25}(r^2+2)\\ =&\,S_{25}-S_9\\ =&\,\frac{25}{6}\\ &\,\hspace{2pt}\times\big(2(25)^2+3(25)\\ &\,\hspace{4pt}+13\big)\\ &\,\hspace{4pt}-\frac96\\ &\,\hspace{6pt}\times\big(2(9)^2+3(9)\\ &\,\hspace{8pt}+13\big)\\ =&\,\frac{25}{6}(1338)\\ &\,\hspace{2pt}-\frac96(202)\\ =&\,5575-303\\ =&\,\boxed{5272}. \end{align*}